Simple Harmonic Motion

$ \begin{aligned} & \text { } \sin ^2 \omega t=\frac{1-\cos 2 \omega t}{2}=\frac{1}{2}-\frac{1}{2} \cos 2 \omega t\\ & \therefore \quad T=\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega} \mathrm{s} \end{aligned} $

According to given figure, springs are connected in parallel.$ \begin{aligned} &\text { Therefore effective spring constant }\\ &\begin{aligned} K_{\mathrm{eff}} & =K_1+K_2 \\ & =K+K=2 K \end{aligned} \end{aligned} $

The restoring force on block due to springs not due to gravity pull

So, $\quad T=2 \pi \sqrt{\frac{M}{K_{\text {eff }}}} \Rightarrow T=2 \pi \sqrt{\frac{M}{2 K}}$

$ \begin{aligned} &\text { Displacement, } x=0.5 \cos (125.6 t)\\ &\begin{aligned} \therefore \quad \omega & =125.6 \mathrm{rad} / \mathrm{s} \\ \therefore \quad T & =\frac{2 \pi}{\omega} \\ & =\frac{2 \times 3.14}{125.6} \approx 0.05 \mathrm{~s} \end{aligned} \end{aligned} $

$A=A_0 e^{-b t}$

Since, $A=50 \%$ of $A_0=\frac{A_0}{2}$ at $t=12 \mathrm{~s}$,

$ \begin{array}{ll} \therefore & \frac{A_0}{2}=A_0 e^{-b t} \\ \Rightarrow & 0.5=e^{-12 b} \end{array} $

$ \begin{aligned} &\text { According to second condition, }\\ &\begin{aligned} & A^{\prime}=A_0 e^{-b \times 36}=A_0\left(e^{-12 b}\right)^3=A_0(0.5)^3 \\ & A^{\prime}=0.125 A_0 \\ & A^{\prime}=\frac{A_0}{8} \\ & \Rightarrow \frac{x}{100} A_0=\frac{A_0}{8} \Rightarrow x=12.5 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \frac{k_1}{k_2}= & \frac{\frac{1}{2} m \omega^2\left(A^2-x_1^2\right)}{\frac{1}{2} m \omega^2\left(A^2-x_2^2\right)} \\ & =\frac{A^2-x_1^2}{A^2-x_2^2}=\frac{A^2-\left(\frac{A}{4}\right)^2}{A^2-\left(\frac{A}{2}\right)^2} \\ & =\frac{\frac{15}{16} A^2}{\frac{3}{4} A^2}=\frac{5}{4} \end{aligned} $

For particle in SHM,

$ \begin{aligned} T & =1.5 \mathrm{~s} \mathrm{so} \\ \omega & =\frac{2 \pi}{T}=\frac{4 \pi}{3} \mathrm{rad} / \mathrm{s} \end{aligned} $

Let oscillations starts from $t=0$ and $x=A \sin \omega t$ is displacement equation Then, KE at time $t$ is,

$ \begin{aligned} K & =\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{d x}{d t}\right)^2 \\ & =\frac{1}{2} m \cdot A^2 \omega^2 \cos ^2 \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

And total energy of particle is,

$ E=\frac{1}{2} k A^2 $

Here, $k=m \omega^2$

$ =\frac{1}{2} m A^2 \omega^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} &\text { Given, ratio, } \frac{K}{E}=\frac{3}{4}\\ &\begin{aligned} & \Rightarrow \quad \frac{\frac{1}{2} m A^2 \omega^2 \cos ^2 \omega t}{\frac{1}{2} m A^2 \omega^2}=\frac{3}{4} \\ & \Rightarrow \quad \cos ^2 \omega t=\frac{3}{4} \\ & \Rightarrow \quad \cos \omega t=\sqrt{\frac{3}{2}} \\ & \Rightarrow \quad \omega t=\frac{\pi}{6} \\ & \Rightarrow \quad t=\frac{\pi}{6 \omega}=\frac{\pi \times 3}{6 \times 4 \pi}=\frac{1}{8} \mathrm{~s} \end{aligned} \end{aligned} $

$ \text {} \begin{aligned} y_1 & =0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right) \\ \therefore \quad v_1 & =0.1 \times 100 \pi \cos \left(100 \pi t+\frac{\pi}{3}\right) \\ & =10 \pi \times \cos \left(100 \pi t+\frac{\pi}{3}\right) \\ y_2 & =0.1 \cos \pi t \\ & =0.1 \sin \left(\pi t+\frac{\pi}{2}\right) \\ \therefore \quad v_2 & =0.1 \times \pi \cos \left(\pi t+\frac{\pi}{2}\right) \\ & =0.1 \pi \cos \left(\pi t+\frac{\pi}{2}\right) \end{aligned} $

$ \begin{aligned} \therefore \text { At } t & =0 \\ v_1 & =10 \cos \frac{\pi}{3} \\ \text { and } v_2 & =0.1 \pi \cos \frac{\pi}{2} \end{aligned} $

$\therefore$ Phase difference between $v_1$ and $v_2$ at time $t=0$,

is given as,

$ \phi=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6} $

Step 1: Find Spring Constant ($k$)

The spring stretches by 0.2 m when a 0.5 kg mass hangs from it.

This means the force from the weight ($m_1 g$) equals the force from the spring ($kx$):

$k x = m_1 g$

So,

$k = \frac{m_1 g}{x} = \frac{0.5 \times 10}{0.2} = 25~\mathrm{N}/\mathrm{m}$

Step 2: Use the New Mass

The new hanging mass is $m_2 = 0.25~\mathrm{kg}$.

Step 3: Find the New Time Period ($T$)

The time period of a mass-spring system is:

$T = 2\pi \sqrt{\frac{m_2}{k}}$

Substitute the values:

$T = 2\pi \sqrt{\frac{0.25}{25}}$

$T = 2\pi \sqrt{0.01}$

$T = 2\pi \times 0.1 = 0.2\pi~\mathrm{s}$

Step 4: Calculate the Value

Using $\pi \approx 3.14$:

$T = 0.2 \times 3.14 = 0.628~\mathrm{s}$

• Mass $ m = 4 \, \text{kg} $

• Force constant (spring constant) $ k = 64 \, \text{N/m} $

Formula for time period of a mass–spring system:

$ T = 2\pi \sqrt{\frac{m}{k}} $

Substitute the values:

$ T = 2\pi \sqrt{\frac{4}{64}} = 2\pi \sqrt{\frac{1}{16}} = 2\pi \times \frac{1}{4} = \frac{\pi}{2} $

✅ Answer: $ \boxed{\frac{\pi}{2}\ \text{s}} $

Correct Option: B

Let angular frequency of SHM is $\omega$ and when is at distance $x$ from mean position, its velocity will be given as

$ v=\omega \sqrt{A^2-x^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

When its velocity becomes double, let new amplitude is $A^{\prime}$.

So, $\quad 2 v=\omega \sqrt{A^{\prime 2}-x^2}$

Putting the value of $v$ from equation (i)

$ \begin{aligned} & 2 \omega \sqrt{A^2-x^2}=\omega \sqrt{A^{\prime 2}-x^2} \\ & 2 \sqrt{A^2-x^2}=\sqrt{A^{\prime 2}-x^2} \\ \Rightarrow \quad & 4\left(A^2-x^2\right)=A^{\prime 2}-x^2 \\ \Rightarrow \quad & A^{\prime 2}=4 A^2-3 x^2 \Rightarrow A^{\prime}=\sqrt{4 A^2-3 x^2} \end{aligned} $

$x=A \sin (\omega t+\theta)$

For maximum displacement, $x=A$

$ \begin{array}{ll} & A=A \sin (\omega t+\theta) \\ \Rightarrow & 1=\sin (\omega t+\theta) \\ \Rightarrow & \omega t+\theta=\frac{\pi}{2} \\ \Rightarrow & \omega t=\frac{\pi}{2}-\theta \end{array} $

$ \Rightarrow \quad t=\frac{\pi}{2 \omega}-\frac{\theta}{\omega} $

Given:

• Maximum velocity $ v_{\text{max}} = 5 \, \text{m/s} $

• Maximum acceleration $ a_{\text{max}} = 10 \, \text{m/s}^2 $

For SHM relationships:

$ v_{\text{max}} = \omega A $

$ a_{\text{max}} = \omega^2 A $

where $ \omega $ is the angular frequency, and $ A $ is the amplitude.

Step 1: Divide the two equations

$ \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{\omega^2 A}{\omega A} = \omega $

$ \Rightarrow \omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{10}{5} = 2 \, \text{rad/s} $

Step 2: Find the time period

$ T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \, \text{s} $

✅ Final Answer:

$ \boxed{T = \pi \, \text{s}} $

Correct Option: A) $ \pi \text{ s} $

Combined mass, $m=m_1+m_2$

$ =1+0.5=1.5 \mathrm{~kg} $

Using conservation of momentum,

$ \begin{aligned} & m_1 u_1+m_2 u_2=\left(m_1+m_2\right) v \\ \Rightarrow \quad & 1 \times 0+0.5 \times 3=1.5 v \\ \Rightarrow \quad & v=1 \mathrm{~m} / \mathrm{s} \end{aligned} $

According to conservation of energy,

$ \begin{aligned} & & \frac{1}{2} m v^2 & =\frac{1}{2} k A^2 \\ \Rightarrow & & A & =\sqrt{\frac{m v^2}{k}} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad \sqrt{\frac{1.5 \times 1^2}{600}} & =\sqrt{\frac{1}{400}}=\frac{1}{20} \mathrm{~m} \\ & =\frac{100}{20} \mathrm{~cm}=5 \mathrm{~cm} \end{aligned} $

$ \text { } \begin{aligned} \frac{K}{U} & =\frac{\frac{1}{2} m \omega^2\left(A^2-x^2\right)}{\frac{1}{2} m \omega^2 x^2} \\ & =\frac{A^2-x^2}{x^2} \\ & =\frac{A^2-\left(\frac{A}{2}\right)^2}{\left(\frac{A}{2}\right)^2}\left[\text { Here, } x=\frac{A}{2}\right] \\ & =\frac{\frac{3 A^2}{4}}{\frac{A^2}{4}}=\frac{3}{1} \\ \therefore K: U & =3: 1 \end{aligned} $

$ \begin{aligned} \text { } \quad & T=2 \pi \sqrt{\frac{m}{k}} \\ \frac{T_1}{T_2} & =\sqrt{\frac{m_1}{m_2}} \\ \frac{T}{T+0.2 \pi} & =\sqrt{\frac{4}{9}}=\frac{2}{3} \\ \Rightarrow \quad 3 T & =2 T+0.4 \pi \\ \Rightarrow \quad T & =0.4 \pi \mathrm{~s} \end{aligned} $

$\therefore$ Using Eq. (i),

$ \begin{aligned} & & T & =2 \pi \sqrt{\frac{m}{k}} \\ & \Rightarrow & 0.4 \pi & =2 \pi \sqrt{\frac{4}{k}} \Rightarrow 0.2=\frac{2}{\sqrt{k}} \\ & \Rightarrow & \sqrt{k} & =\frac{1}{0.1} \Rightarrow \sqrt{k}=10 \\ & \Rightarrow & k & =100 \mathrm{Nm}^{-1} \end{aligned} $

Total energy,

$ \begin{aligned} & E=\frac{1}{2} m \omega^2 A^2 \\ & K+U=\frac{1}{2} m \omega^2(0.05)^2 \\ \Rightarrow \quad & 4 \times 10^{-3}+\frac{1}{2} m \omega^2 x^2=\frac{1}{2} m \omega^2(0.05)^2 \\ \Rightarrow \quad & \frac{1}{2} m \omega^2\left[(0.05)^2-x^2\right]=4 \times 10^{-3} \\ \Rightarrow \quad & \frac{1}{2} m \omega^2\left[(0.05)^2-(0.03)^2\right]=4 \times 10^{-3} \\ \Rightarrow \quad & \frac{1}{2} m \omega^2=2.5 \\ \Rightarrow \quad & m \omega^2=5 \end{aligned} $

$\therefore$ Maximum force

$ F_{\max }=m \omega^2 A=5 \times 0.05=0.25 \mathrm{~N} $

$ \begin{aligned} &\text { Angular velocity of the system, }\\ &\begin{aligned} \omega & =\sqrt{\frac{k}{m_{\text {total }}}}=\sqrt{\frac{600}{(1+0.5)}} \\ & =\sqrt{\frac{600}{1.5}}=20 \mathrm{rad} / \mathrm{s} \\ \therefore \quad f & =\frac{\omega}{2 \pi}=\frac{20}{2 \pi}=\frac{10}{\pi} \mathrm{~Hz} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y=5 \sin (3 \pi t)+5 \sqrt{3} \cos (3 \pi t) \\ = & 10\left[\frac{5}{10} \sin 3 \pi t+\frac{5 \sqrt{3}}{10} \cos (3 \pi t)\right] \\ = & 10\left[\sin 3 \pi t \cdot\left(\frac{1}{2}\right)+\cos 3 \pi t \cdot\left(\frac{\sqrt{3}}{2}\right)\right] \\ = & 10\left[\sin (3 \pi t) \cos \frac{\pi}{3}+\cos 3 \pi t \sin \frac{\pi}{3}\right] \\ = & 10 \sin \left(3 \pi t+\frac{\pi}{3}\right) \\ \therefore & A=10 \mathrm{~cm} \end{aligned} \end{aligned} $

$ \begin{aligned} & n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{25}{0.1}} \\ & n=\frac{5}{2 \pi} \mathrm{~Hz} \end{aligned} $

$\therefore$ Angular frequency, $\omega=2 \pi n$

$ =2 \pi \times \frac{5}{2 \pi}=5 \mathrm{rad} \mathrm{~s}^{-1} $

To determine the frequency of oscillation for the system, we need to consider the concept of reduced mass, which is useful when analyzing systems involving two masses connected by a spring.

First, the reduced mass ($ \mu $) of a two-body system is given by the following formula:

$ \frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2} $

where $ m_1 $ and $ m_2 $ are the masses of the two blocks connected to the spring. The reduced mass essentially simplifies the system to behave as if it were a single object of mass $\mu$.

Once the reduced mass is calculated, the frequency of oscillation ($ \omega $) for a spring-mass system is determined by the formula:

$ \omega = \sqrt{\frac{k}{\mu}} $

Substituting the expression for reduced mass into the formula, we get:

$ \omega = \sqrt{k \left(\frac{1}{m_1} + \frac{1}{m_2}\right)} $

This expression gives the angular frequency of the simple harmonic motion for the system, where $ k $ is the spring constant. This encapsulates how the two masses and the spring constant interact to influence the oscillation frequency.

To find the ratio of amplitudes, $\frac{A_1}{A_2}$, for the simple harmonic motion (SHM) described, we need to consider the changes when a mass $m$ is added to a mass $M$, which is attached to a spring.

Key Equations:

Time Period for Mass $(M + m)$:

$ T_2 = 2 \pi \sqrt{\frac{m + M}{k}} $

Time Period for Mass $M$ Only:

$ T_1 = 2 \pi \sqrt{\frac{M}{k}} $

Velocity Considerations:

Assume $v_1$ is the velocity of the mass $M$ at the mean position and $v_2$ is the velocity of the combined mass $(m+M)$.

Using Conservation of Linear Momentum:

$ M v_1 = (m + M) v_2 $

Since $v_1 = A_1 \omega_1$ and $v_2 = A_2 \omega_2$:

$ M (A_1 \omega_1) = (m + M) (A_2 \omega_2) $

Rewriting for the amplitude ratio $\frac{A_1}{A_2}$:

$ \frac{A_1}{A_2} = \frac{(m + M)}{M} \times \frac{\omega_2}{\omega_1} $

Given $\omega_1 = \frac{2 \pi}{T_1}$ and $\omega_2 = \frac{2 \pi}{T_2}$, we have:

$ \frac{\omega_2}{\omega_1} = \frac{T_1}{T_2} $

Substituting Time Periods:

Substituting from earlier equations:

$ \frac{T_1}{T_2} = \sqrt{\frac{M}{m + M}} $

Therefore:

$ \frac{A_1}{A_2} = \frac{(m + M)}{M} \times \sqrt{\frac{M}{m + M}} $

$ \frac{A_1}{A_2} = \sqrt{\frac{m + M}{M}} $

This provides the required ratio of the amplitudes when a smaller mass is attached, leading to a new simple harmonic motion with the described properties.

The particle has a mass of:

$ m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} $

The displacement of the particle is described by:

$ x = 8 \cos \left(50 t + \frac{\pi}{12}\right) \, \text{m} $

The maximum velocity $ v_{\text{max}} $ can be determined using the formula:

$ v_{\text{max}} = A \omega $

where $ A = 8 \, \text{m} $ and $ \omega = 50 \, \text{rad/s} $. Thus:

$ v_{\text{max}} = 8 \times 50 = 400 \, \text{m/s} $

The maximum kinetic energy ($\text{KE}_{\text{max}}$) is given by:

$ \text{KE}_{\text{max}} = \frac{1}{2} m \left(v_{\text{max}}\right)^2 $

Substituting the known values:

$ \text{KE}_{\text{max}} = \frac{1}{2} \times 2 \times 10^{-3} \times \left(4 \times 10^2\right)^2 $

$ = 10^{-3} \times 16 \times 10^4 $

Therefore, the maximum kinetic energy is:

$ \text{KE}_{\text{max}} = 160 \, \text{J} $

Given:

Mass of the particle: $ m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} $

Displacement-force relationship: $ 500F + \pi^2 y = 0 $

From this equation, we can express the force $ F $ in terms of the displacement $ y $:

$ F = \frac{-\pi^2 y}{500} $

In simple harmonic motion (SHM), force is also given by:

$ F = -ky $

By comparing the two force equations, we identify the spring constant $ k $:

$ k = \frac{\pi^2}{500} $

The angular frequency $ \omega $ for SHM is determined by:

$ \omega = \sqrt{\frac{k}{m}} $

Substituting the values for $ k $ and $ m $:

$ \omega = \sqrt{\frac{\pi^2}{500 \times 2 \times 10^{-3}}} = \pi \, \text{rad/s} $

The time period $ T $ of SHM is given by:

$ T = \frac{2\pi}{\omega} $

Plugging in the value of $ \omega $:

$ T = \frac{2\pi}{\pi} \Rightarrow T = 2 \, \text{seconds} $

Therefore, the time period of oscillation of the particle is 2 seconds.

Here, given

$ \begin{aligned} & \begin{aligned} y_1 & =5[\sin 2 \pi t+\sqrt{3} \cos 2 \pi t] \\ y_2 & =5 \sin \left[2 \pi t+\frac{\pi}{4}\right] \\ \text { For } y_1 & =10\left[\frac{1}{2} \sin 2 \pi t+\frac{\sqrt{3}}{2} \cos 2 \pi t\right] \\ & =10\left(\cos \frac{\pi}{3} \sin 2 \pi t+\sin \frac{\pi}{3} \cos 2 \pi t\right) \\ y_1 & =10\left[\sin 2 \pi t+\frac{\pi}{3}\right] \end{aligned} \end{aligned} $

Amplitude in $y_1$ is $A_1=10$

Similarly, $y_2=5 \sin \left(2 \pi t+\frac{\pi}{4}\right)$

Amplitude in $y_2$ is $A_2=5$

Thus, $\frac{A_1}{A_2}=\frac{10}{5}=\frac{2}{1}$

When the mass is connected to the two spring individually.

$ \begin{aligned} & v_1=\frac{1}{2 \pi} \sqrt{\frac{k_1}{m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & v_2=\frac{1}{2 \pi} \sqrt{\frac{k_2}{m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Now, the block is connected with two springs, $k_{\text {eq }}=k_1+k_2$

Time period, $T=2 \pi \sqrt{\frac{m}{k_{\text {eq }}}}$

$ =2 \pi \sqrt{\frac{m}{k_1+k_2}} $

$ \begin{aligned} \text { Frequency } & =\frac{1}{T}=\frac{1}{2 \pi} \times \sqrt{\frac{k_1+k_2}{m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \\ v & =\frac{1}{2 \pi}\left[\frac{k_1}{m}+\frac{k_2}{m}\right]^{1 / 2} \end{aligned} $

$ \begin{aligned} &\text { From Eq. (i) } \frac{k_1}{m}=4 \pi^2 v_1^2 \text { and } \frac{k_2}{m}=4 \pi v_2^2\\ \,\,\,&\begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,v & =\frac{1}{2 \pi}\left[\frac{4 \pi^2 v_1^2}{1}+\frac{4 \pi^2 v_2^2}{1}\right]^{1 / 2} \\ & =\frac{2 \pi}{2 \pi}\left[v_1^2+v_2^2\right]^{1 / 2} \\ v & =\sqrt{v_1^2+v_2^2} \end{aligned} \end{aligned} $

We start with the formula for the time period of a magnetic dipole in a magnetic field:

$ T = 2\pi \sqrt{\frac{I}{MB}} \quad \text{...(i)} $

where $ I $ is the moment of inertia, $ M $ is the magnetic moment, and $ B $ is the magnetic field.

Since the moment of inertia $ I $ is directly proportional to the mass $ m $ of the magnet—meaning $ I \propto m $—if the mass is increased by 9 times, then:

$ I' = 9I $

The new time period $ T' $ can be expressed as:

$ T' = 2\pi \sqrt{\frac{9I}{MB}} \quad \text{...(ii)} $

By dividing equation (ii) by equation (i), we find:

$ \frac{T'}{T} = \frac{2\pi \sqrt{\frac{9I}{MB}}}{2\pi \sqrt{\frac{I}{MB}}} $

Simplifying gives:

$ \frac{T'}{T} = \sqrt{9} = 3 $

Therefore, the new time period $ T' $ is three times the original time period:

$ T' = 3T $

Given, $k=9 \pi^2 \mathrm{~N} / \mathrm{m}$

Reduced mass of system is

$ \begin{aligned} \mu & =\frac{m_1 m_2}{m_1+m_2} \\ \Rightarrow \quad \mu & =\frac{3 \times 3}{3+3}=\frac{3}{2} \mathrm{~kg} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, parallel combination of spring constant (between (1) and (2))

$ k_{\mathrm{eq}}=k+k=2 k $

Now, $k_{\text {eq }}$ and (3) are in series connection,

$ \begin{aligned} k_{\text {eq }}^{\prime} & =\frac{2 k \times k}{2 k+k}=\frac{2}{3} k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & =\frac{2}{3} \times 9 \pi^2 \\ k_{\text {eq }}^{\prime} & =6 \pi^2 \end{aligned} $

Time period is given by formula,

$ \begin{aligned} T & =2 \pi \sqrt{\frac{\mu}{k}} \\ & =2 \pi \sqrt{\frac{3}{2} \times \frac{1}{6 \pi^2}}=1 \mathrm{~s} \end{aligned} $

Given the scenario where a body is undergoing simple harmonic motion:

The potential energy at displacement $x$ is denoted as $E_1$.

Since the body is in simple harmonic motion, the relation is given by:

$ E_1 = \frac{1}{2} m \omega^2 x^2 $

From this, we can derive:

$ \sqrt{E_1} = x \sqrt{\frac{1}{2} m \omega^2} \quad \text{...(i)} $

At displacement $y$, the potential energy is $E_2$.

Similarly:

$ E_2 = \frac{1}{2} m \omega^2 y^2 $

Leading to:

$ \sqrt{E_2} = y \sqrt{\frac{1}{2} m \omega^2} \quad \text{...(ii)} $

For displacement $(x+y)$, the potential energy is denoted by $E$.

It can be expressed as:

$ E = \frac{1}{2} m \omega^2 (x+y)^2 $

Therefore:

$ \sqrt{E} = (x+y) \sqrt{\frac{1}{2} m \omega^2} = x \sqrt{\frac{1}{2} m \omega^2} + y \sqrt{\frac{1}{2} m \omega^2} $

By using equations (i) and (ii), we find:

$ \sqrt{E} = \sqrt{E_1} + \sqrt{E_2} $

Given:

Mass, $ m = 1 \, \text{kg} $

Displacement, $ x = 0.2 \, \text{m} $

Period, $ T = \frac{\pi}{2} $

We need to find the potential energy at this displacement.

First, we recall the formula for the period of oscillation:

$ T = \frac{2\pi}{\omega} $

The angular frequency $\omega$ is related to the mass $ m $ and the spring constant $ k $ by:

$ \omega = \sqrt{\frac{k}{m}} $

Substituting this expression for $\omega$ into the period equation, we have:

$ T = \frac{2\pi \cdot \sqrt{m}}{\sqrt{k}} $

Given $ T = \frac{\pi}{2} $, we can solve for $ k $:

$ \frac{\pi}{2} = \frac{2\pi \times \sqrt{1}}{\sqrt{k}} $

Solving for $ k $:

$ \frac{\pi}{2} = \frac{2\pi}{\sqrt{k}} $

$ \sqrt{k} = 4 $

$ k = 16 \, \text{N/m} $

Now, compute the potential energy $ U $ at the displacement $ x = 0.2 \, \text{m} $ using:

$ U = \frac{1}{2} k x^2 $

Substitute the values:

$ U = \frac{1}{2} \times 16 \times (0.2)^2 $

$ U = 8 \times 0.04 $

$ U = 0.32 \, \text{J} $

Therefore, the potential energy at a displacement of 0.2 m is $ 0.32 \, \text{J} $.

Given a particle with mass $ m = 10 \, \text{g} = 0.01 \, \text{kg} $, the potential energy as a function of displacement $ x $ is expressed as:

$ U(x) = 50x^2 + 100 $

We recognize that the potential energy for a harmonic oscillator is given by:

$ U(x) = \frac{1}{2} k x^2 $

To find the spring constant $ k $, we compare the two expressions:

$ \frac{1}{2} k x^2 = 50 x^2 $

Solving for $ k $:

$ \frac{1}{2} k = 50 \\ k = 100 $

The frequency of oscillation $ f $ is derived from the angular frequency $\omega$, given by:

$ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \cdot \sqrt{\frac{k}{m}} $

Substituting the values for $ k $ and $ m $:

$ f = \frac{1}{2\pi} \cdot \sqrt{\frac{100}{0.01}} \\ f = \frac{100}{2\pi} \\ f = \frac{50}{\pi} \, \text{s}^{-1} $

To find the minimum coefficient of friction ($\mu$) required for a heavy body to remain stationary on a horizontally oscillating board, we start with the given amplitude and period:

Amplitude ($A$) = 0.3 m

Period ($T$) = 4 s

The force associated with an oscillating system is given by $F = kx$. For the body not to slip due to the board's oscillations, the frictional force ($f$) must be equal to the force due to the oscillation. Thus, we have:

$ f = \mu N = \mu mg $

Replacing the spring force expression with terms for mass ($m$) and angular frequency ($\omega$), we have:

$ \mu mg = m\omega^2 A $

Simplifying this expression to solve for $\mu$:

$ \mu = \frac{\omega^2 A}{g} $

Knowing that the angular frequency $\omega$ can be expressed in terms of the period:

$ \omega = \frac{2\pi}{T} $

We substitute $\omega$ to find $\mu$:

$ \mu = \left(\frac{2\pi}{T}\right)^2 \frac{A}{g} $

Substituting the provided values:

$ \mu = \left(\frac{2\pi}{4}\right)^2 \frac{0.3}{9.8} $

Calculate $\mu$:

$ \mu = \frac{4 \times \pi^2 \times 0.3}{4 \times 4 \times 9.8} $

Finally, compute the numerical value:

$ \mu = 0.075 $

Thus, the minimum coefficient of friction required is $\mu = 0.075$.

The system consists of a test tube and mercury. The total mass is calculated as:

$ \text{Total mass (tube + mercury)} = 6 \, \text{g} + 10 \, \text{g} = 16 \, \text{g} = 0.016 \, \text{kg} $

The area of the test tube's cross-section is given as:

$ A = 10 \, \text{cm}^2 = 10^{-3} \, \text{m}^2 $

The density of water is:

$ \rho = 10^3 \, \text{kg/m}^3 $

The effective spring constant $ k $ due to buoyancy is determined using the formula:

$ k = A \times \rho \times g $

Substituting the known values:

$ k = 10^{-3} \times 10^3 \times 10 = 10 $

Thus, the spring constant $ k $ is 10.

The time period of oscillation $ T $ is given by:

$ T = 2 \pi \sqrt{\frac{m}{k}} $

Substituting the mass $ m = 0.016 \, \text{kg} $ and spring constant $ k = 10 $:

$ T = 2 \pi \sqrt{\frac{0.016}{10}} = 0.25 \, \text{s} $

Therefore, the time period of oscillation is 0.25 seconds.

To determine the acceleration of the block at its lowest position, let's start by examining the system's mechanics.

1. Maximum Displacement Calculation

Let $ x $ be the maximum downward displacement of the block. The energy conservation equation, considering the force exerted by the springs and gravitational potential energy, is:

$ mgx = \frac{1}{2}(k_1 + k_2)x^2 $

Substitute the known values:

$ x = \frac{2mg}{k_1 + k_2} = \frac{2 \times 3 \times 10}{50 + 150} = \frac{60}{200} = 0.3 \, \text{m} $

2. Acceleration Calculation at Maximum Displacement

The acceleration of the block at this position, due to the net force from the springs and gravity, is given by:

$ a = \frac{(k_1 + k_2)x - mg}{m} $

Plug in the calculated values:

$ a = \frac{200 \times 0.3 - 3 \times 10}{3} = \frac{60 - 30}{3} = \frac{30}{3} = 10 \, \text{m/s}^2 $

Thus, the acceleration of the block at its lowest point is $ 10 \, \text{m/s}^2 $.

The amplitude of a damped oscillator decreases over time due to the damping effect. The relationship for the amplitude is given by:

$ A' = A e^{-\frac{bt}{2m}} $

where $ A $ is the initial amplitude, $ b $ is the damping coefficient, $ t $ is the time, and $ m $ is the mass of the oscillator.

Initially, at $ t = 2 $ seconds, the amplitude becomes $\frac{1}{e}$ of the initial amplitude:

$ \frac{A}{e} = A e^{-\frac{b \times 2}{2m}} $

This simplifies to:

$ \frac{A}{e} = \frac{A}{e^{\frac{b}{m}}} $

From this, we deduce:

$ \frac{b}{m} = 1 $

At $ t = 4 $ seconds (2 more seconds), the amplitude further becomes:

$ A' = A e^{-\frac{b \times 4}{2m}} = \frac{A}{e^{\frac{4b}{2m}}} = \frac{A}{e^2} $

Thus, the amplitude at $ t = 4 $ seconds is proportional to $\frac{1}{e^2}$.

Given that the time period of the particle executing simple harmonic motion (SHM) is $ T = 3 \, \text{s} $.

From this, we know:

$ T = \frac{2\pi}{\omega} \Rightarrow \omega = \frac{2\pi}{3} $

Let $ A $ represent the maximum amplitude. The displacement of the particle is 60% of its amplitude, which means:

$ y = 0.6A $

For SHM, the kinetic energy (KE) is given by:

$ \mathrm{KE} = \frac{1}{2} m \omega^2 \left(A^2 - y^2\right) $

The potential energy (PE) is:

$ \mathrm{PE} = \frac{1}{2} m \omega^2 y^2 $

The ratio of kinetic energy to potential energy is:

$ \frac{\mathrm{KE}}{\mathrm{PE}} = \frac{\frac{1}{2} m \omega^2 \left(A^2 - y^2\right)}{\frac{1}{2} m \omega^2 y^2} = \frac{A^2 - y^2}{y^2} $

Substituting $ y = 0.6A $:

$ \begin{aligned} \frac{\mathrm{KE}}{\mathrm{PE}} &= \frac{A^2 - (0.6A)^2}{(0.6A)^2} \\ &= \frac{A^2(1 - 0.36)}{0.36A^2} \\ &= \frac{0.64A^2}{0.36A^2} \\ &= \frac{0.64}{0.36} \\ &= \frac{16}{9} \end{aligned} $

Thus, the ratio of the kinetic to potential energy when the displacement is 60% of the amplitude is $\frac{16}{9}$.

The displacement of a particle undergoing simple harmonic motion is given by $ y = A \sin (2t + \phi) $, where $ t $ represents time in seconds, and $ \phi $ is the phase angle. At $ t = 0 $, the displacement is 2 m, and the velocity is $ 4 \, \text{ms}^{-1} $.

Displacement Equation:

$ 2 = A \sin (2t + \phi) \quad \text{(Equation 1)} $

Velocity Equation:

The velocity $ v $ is the derivative of displacement with respect to time:

$ v = \frac{dy}{dt} = 2A \cos (2t + \phi) $

At $ t = 0 $:

$ 4 = 2A \cos (2t + \phi) $

$ \Rightarrow 2 = A \cos (2t + \phi) \quad \text{(Equation 2)} $

Equating Equations (1) and (2):

$ A \sin (2t + \phi) = A \cos (2t + \phi) $

Simplifying gives:

$ \sin (2t + \phi) = \cos (2t + \phi) $

$ \sin (2t + \phi) = \sin \left[90^\circ - (2t + \phi)\right] $

This implies:

$ 2t + \phi = 90^\circ - (2t + \phi) $

$ 4t + 2\phi = 90^\circ $

$ 2t + \phi = 45^\circ $

When $ t = 0 $:

$ \phi = 45^\circ $

Therefore, the phase angle $\phi$ is $ 45^\circ $.

To find the time required for the amplitude of a damped oscillator to become $\frac{1}{e^{1.2}}$ times its initial value, consider the displacement of the oscillator given by:

$ x(t) = \exp(-0.2t) \cos(3.2t + \phi) $

Amplitude Analysis

The amplitude of the oscillator is represented as:

$ A = e^{-0.2t} $

Initial Amplitude at $ t = 0 $:

At $ t = 0 $, the amplitude $ A_1 $ is:

$ A_1 = e^{-0.2 \times 0} = 1 $

Amplitude at $ t = t' $:

At time $ t = t' $, the amplitude $ A_2 $ should be:

$ A_2 = \frac{1}{e^{1.2}} A_1 = \frac{1}{e^{1.2}} \times 1 = \frac{1}{e^{1.2}} $

Solving for $ t' $

To find $ t' $ such that:

$ \frac{1}{e^{1.2}} = e^{-0.2t} $

This relation simplifies to:

$ e^{1.2} = e^{0.2t} $

Thus, we have:

$ 1.2 = 0.2t $

Therefore, solving for $ t $:

$ t = \frac{1.2}{0.2} = 6 \text{ seconds} $

Hence, the time required for the amplitude to reduce to $\frac{1}{e^{1.2}}$ times its initial value is 6 seconds.

Given:

Time period of oscillation of a pendulum in air is $ T $.

Density of the bob, $\rho = \frac{5000}{3} \, \text{kg/m}^3 $.

Density of water, $\sigma = 1000 \, \text{kg/m}^3 $.

We need to find the ratio $\frac{T}{t}$, where $ t $ is the time period of the pendulum's oscillation in water.

The time period of oscillation of a pendulum in water can be expressed as:

$ t = \frac{T}{\sqrt{1 - \frac{\sigma}{\rho}}} $

Substitute the given values:

$ t = \frac{T}{\sqrt{1 - \frac{1000 \times 3}{5000}}} $

This simplifies to:

$ t = \frac{T}{\sqrt{1 - \frac{3000}{5000}}} $

Simplifying further:

$ t = \frac{T}{\sqrt{1 - 0.6}} $

$ t = \frac{T}{\sqrt{0.4}} $

The ratio $\frac{T}{t}$ is:

$ \frac{T}{t} = \sqrt{\frac{1}{0.4}} $

$ \frac{T}{t} = \sqrt{\frac{5}{2}} $

Thus, the value of $\frac{T}{t}$ is $\sqrt{\frac{5}{2}}$.

For a body performing simple harmonic motion, the relation between velocity and displacement

$ v=\omega \sqrt{A^2-x^2} $

Square both side

$ v^2=\omega^2\left(A^2-x^2\right) $

For ( $\omega=1$ )

$ v^2+x^2=A^2 $

(a) So, the graph between velocity and displacement will be a circle.

(b) We know that,

$ x=A \sin \omega t \quad \ldots \text { (i) }$

Differentiate w.r.t. to $t$

$ \frac{d x}{d t}=v=+\omega A \cos \omega t \quad \ldots \text { (ii) }$

Again differentiate w.r.t. to $t$,

$ \frac{d^2 x}{d t^2}=a=-\omega^2 A \sin \omega t \quad \ldots \text { (iii) }$

From Eq. (i),

$ a=-\omega^2 x \quad \ldots \text { (iv) }$

Comparing from Eq. $y=m x+c$

It is a equation of straight line with slope, $m=-\omega^2$.

So, the graph between acceleration and displacement will be a straight line.

(c) From Eq. (iii),

$ \frac{d^2 x}{d t^2}=a=-\omega^2 A \sin \omega t $

The graph between acceleration and time will be a function of sine wave.

(d) From Eq. ( $A$ ),

$ \begin{aligned} & v=\omega \sqrt{A^2-\omega^2} \\ & v^2=\omega^2\left(A^2-x^2\right) \\ & x^2=\left(-\frac{v^2}{\omega^2}+A^2\right) \\ & x=\sqrt{A^2-\frac{v^2}{\omega^2}} \end{aligned} $

Putting this value in Eq. (iv),

$ a=-\omega^2 \sqrt{A^2-\frac{v^2}{\omega^2}} $

Square both side

$ \begin{aligned} a^2 & =\omega^4\left(A^2-\frac{v^2}{\omega^2}\right) \\ a^2 & =\omega^4 A^2-v^2 \omega^2 \\ \left(\frac{a}{\omega^2}\right)^2 & =A^2-\frac{v^2}{\omega^2} \\ \left(\frac{a}{\omega^2}\right)^2 & +\left(\frac{v}{\omega}\right)^2=A^2 \end{aligned} $

When $\omega \neq 1$, the graph between acceleration $a$ as a function of velocity $v$ in SHM is an ellipse.

Given, instantaneous displacement of particle executing SHM,

$\begin{aligned} x & =A \sin ^2(\omega t-4) \\ \Rightarrow x & =A\left[\frac{1-\cos 2(\omega t-4)}{2}\right] \\ & \left(\text { Since, } \sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\right) \end{aligned}$

$\Rightarrow x=\frac{A}{2}[1-\cos (2 \omega t-8)]$

Here, angular velocity of particle

$\begin{aligned} & \omega^{\prime}=2 \omega \\ & \Rightarrow \quad \frac{2 \pi}{T^{\prime}}=2 \times \frac{2 \pi}{T} \Rightarrow T^{\prime}=\frac{T}{2} \\ & =\frac{\frac{2 \pi}{\omega}}{2}=\frac{\pi}{\omega} \\ & \left(\therefore T=\frac{2 \pi}{\omega}\right) \\ \end{aligned}$

The mechanical energy lost in each cycle is proportional to the square of the amplitude $ A $. Therefore, the fractional change in energy can be expressed as:

$\begin{aligned} \frac{\Delta E}{E} \simeq \frac{dE}{E} & = \frac{dA^2}{A^2} \\ & = \frac{2AdA}{A^2} = 2 \frac{dA}{A} \\ & = 2 \times 1.5\% = 3\% \end{aligned}$

Thus, the mechanical energy of the oscillator lost in each cycle is 3%.

According to first condition,

In S. H. M, at displacement $x$

Potential energy $=9 \mathrm{~J}$

-i.e, $\frac{1}{2} k x^2=9 \quad \text{... (i)}$

Again, at displacement $y$, Potential energy

$\frac{1}{2} k y^2=16 \quad \text{... (ii)}$

Dividing Eq. (ii) by Eq. (i), we get

$\begin{aligned} \frac{\frac{1}{2} k y^2}{\frac{1}{2} k x^2} & =\frac{16}{9} \\ \Rightarrow \quad\left(\frac{y}{x}\right)^2 & =\left(\frac{4}{3}\right)^2 \\ \frac{y}{x} & =\frac{4}{3} \quad \text{... (iii)} \end{aligned}$

$\because$ Potential energy of body at displacement $(x+y)$

$\begin{aligned} \mathrm{PE} & =\frac{1}{2} k(x+y)^2 \\ & =\frac{9}{x^2}(x+y)^2 \quad \text{[From Eq. (i)]}\\ & =9 \frac{(x+y)^2}{x^2}=9\left(1+\frac{y}{x}\right)^2 \\ & =9\left(1+\frac{4}{3}\right)^2 \quad \text{[From Eq. (iii)]}\\ & =49 \mathrm{~J} \end{aligned}$

If the hydrometer is in equilibrium, the weight is balanced by the buoyancy force on it by the fluid. When hydrometer liquid is pressed by $x$ distance the net unbalanced force in the upward direction is given by $F=-$(Area $\times$ Density $\times$ Gravity $\times$ Distance moved by fluid)

Where, $-v e$ sign shows that the direction of force will be opposite to the direction of motion.

Thus, $m \frac{d^2 x}{d t^2}=-\pi r^2 \rho g x$

Were, $r$ = radius of liquid column

$\rho=$ density of liquid

$x=$ distance moved by liquid

Now, $\quad \frac{d^2 x}{d t^2}=\frac{-\pi r^2 \rho g}{m} x$

or $\quad \omega^2=\frac{\pi r^2 \rho g}{m}$ or $\omega=\sqrt{\frac{\pi r^2 \rho g}{m}}$

$\Rightarrow \quad T=2 \pi \sqrt{\frac{m}{\pi r^2 \rho g}}$

The velocity of the object becomes zero at maximum displacement of either side of the mean position in the SHM.

Time taken from one maximum to another maximum $=0.5 \mathrm{~s}$

Time period $=2 \times 0.5=1 \mathrm{~s}$

Angular velocity, $\omega=2 \pi v$

$=\frac{2 \pi}{T}=\frac{2 \pi}{l}=2 \pi \mathrm{~rad} / \mathrm{s}$

Let initially the cone be in equilibrium i.e. floating

$\begin{aligned} & \text { From the diagram, it is clear that } \\ & \text { Buoyant force }\left(F_b\right)=\text { weight of the cone. }\left(F_w\right) \\ & \qquad \rho_w \times\left(\frac{1}{3} \pi r^2 x\right) \times g=\rho_c \times\left(\frac{1}{3} \pi R^2 H\right) \times g \end{aligned}$

$\Rightarrow \quad 2 r^2 x=R^2 H \quad\left(\because \rho_c=\frac{\rho_w}{2}\right)$

$\Rightarrow \quad x=\frac{R^2 H}{2 r^2} \quad \text{... (i)}$

From the diagram,

$\begin{array}{ll} & \tan \theta=\frac{r}{x}=\frac{R}{H} \\ \Rightarrow \quad & x=\frac{r H}{R} \quad \text{... (ii)} \end{array}$

From Eqs. (i) and (ii)

$\begin{gathered} \frac{R^2 H}{2 r^2}=\frac{r H}{R} \\ \Rightarrow \quad \frac{R^3}{2}=r^3 \Rightarrow r=\left(\frac{1}{2}\right)^{1 / 3} R \end{gathered}$

Now, the cone. is depressed by small distance $\delta(<< H)$ and released it executes simple harmonic motion. In simple harmonic motion, the buoyancy force acts as restoring force which is given as

$F_b=\text { density } \times \text { volume } \times \text { gravity }$

Since, the displacement is very small, the excess submerged shape of cone. can be considered as a cylinder whose volume is given by $\pi r^2 \delta$.

$\therefore \quad F_b=\rho_w \pi r^2 \delta g=\rho_w \pi\left[\left(\frac{1}{2}\right)^{1 / 3} R\right]^2 \delta g$

Thus, force is equal to restoring force, hence

$F_b=k \delta=m \omega^2 \delta$

$\begin{array}{ll} & \rho_w \pi\left(\frac{1}{4}\right)^{1 / 3} R^2 \delta g=\rho_c \frac{1}{3} \pi R^2 H \delta \omega^2 \\ & 2 \times 3 \times\left(\frac{1}{4}\right)^{1 / 3} \delta g=H \delta \omega^2 \quad\left(\because \rho_c=\frac{\rho_w}{2}\right) \\ \Rightarrow \quad & \frac{6\left(\frac{1}{4}\right)^{1 / 3} g}{H}=\omega^2 \\ \text { or } & \omega=\sqrt{\frac{6 g}{H}\left(\frac{1}{4}\right)^{1 / 3}} \Rightarrow 2 \pi f=\sqrt{\frac{6 g}{H}\left(\frac{1}{4}\right)^{1 / 3}} \\ \Rightarrow \quad & f=\frac{1}{2 \pi} \sqrt{\frac{6 g}{H}\left(\frac{1}{4}\right)^{1 / 3}}=\frac{1}{2 \pi} \sqrt{\frac{6 g}{H} \cdot \frac{1}{4^{1 / 3}}} \end{array}$

As we know that, for less damping, resonance peak is taller and sharper.

Let, wave displacement be x.

As we know that,

Potential energy, $U=\frac{1}{2}kx^2$

where, k is wave constant.

Potential energy will be maximum, when displacement is maximum.

$\Rightarrow x=\pm~A$

Let,

Time period $=T$

Length of simple pendulum $=l$

Acceleration due to gravity $=g$

$\because \quad T=2 \pi \sqrt{l / g}$ .... (i)

When sphere is fully filled with water, centre of mass lies at centre.

Case I For, water level below centre of sphere

$l_1 > l_0$ ...... (ii)

Case II For empty sphere,

$l=l_0 < l_1$ ..... (iii)

$\therefore$ From Eqs. (i), (ii) and (iii)

In case I, time period increase.

In case II, time period decrease.

Given, mass of block, $m=1 \mathrm{~kg}$

Spring constant, $k=100 \mathrm{~N} / \mathrm{m}$

In case I, extension in spring, $\Delta x_1=10 \mathrm{~cm}$

$=10 \times 10^{-2} \mathrm{~m}$

Initial speed, $u=0 \mathrm{~ms}^{-1}$

In case II, extension in spring $\Delta x_2=5 \mathrm{~cm}$

$=5 \times 10^{-2} \mathrm{~m}$

Let, kinetic energy and potential energy be (KE) and (PE)

As we know that,

$\begin{aligned} & \mathrm{KE}=\frac{1}{2} k\left(\Delta x_1^2-\Delta x_2^2\right) \\ &=\frac{1}{2} \times 100\left(10^2-5^2\right) \times 10^{-4} \\ &=50 \times 75 \times 10^{-4}=0.375 \mathrm{~J} \\ & \text { and } \mathrm{PE}=\frac{1}{2} k \Delta x_2^2 \\ &=\frac{1}{2} \times 100 \times\left(5 \times 10^{-2}\right)^2=50 \times 25 \times 10^{-4}=0.125 \mathrm{~J} \end{aligned}$

Given, range of mass is from $m_1$ to $m_2$ i.e. 0 to $15 \mathrm{~kg}$

Length of spring, $x=0.25 \mathrm{~m}$

Time period of oscillation, $T=\frac{2 \pi}{5} \mathrm{~s}$

Let mass of body suspended $=m$

Spring constant $=k$

Acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$

Since, $T =2 \pi \sqrt{\frac{m}{k}}$ ..... (i)

$\begin{aligned} \text{and} \quad F & =m g=k x \\ \Rightarrow \quad k & =\frac{m g}{x}=\frac{15 \times 10}{0.25} \mathrm{~N} / \mathrm{m} \end{aligned}$

Substituting in Eq. (i), we get

$\begin{aligned} & T=2 \pi \sqrt{\frac{m \times 25}{15 \times 10 \times 100}} \\ & \Rightarrow \quad\left(\frac{2 \pi}{5}\right)^2=(2 \pi)^2\left(\frac{m}{15 \times 40}\right) \\ & \Rightarrow \quad \frac{4 \pi^2}{25}=\frac{4 \pi^2 m}{15 \times 40} \Rightarrow m=24 \mathrm{~kg} \\ \end{aligned}$