Simple Harmonic Motion

$ \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} $

$ \begin{aligned} &\text { For the given spring mass system, }\\ &k_{\mathrm{eq}}=k+2 k+k=4 k \end{aligned} $

$ \text { ∴ Frequency, } \eta=\frac{1}{2 \pi} \sqrt{\frac{k_{\text {eq }}}{m}}=\frac{1}{2 \pi} \sqrt{\frac{4 k}{m}} $

The time period of a simple pendulum is given by the formula:

$T = 2\pi \sqrt{\frac{l}{g}}$

where $T$ is the time period, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity at the location of the pendulum.

The acceleration due to gravity changes with height above the Earth's surface. The acceleration due to gravity at a height $h$ above the Earth's surface can be expressed as:

$g' = g \left(\frac{R}{R + h}\right)^2$

where $g$ is the acceleration due to gravity at the surface of the Earth, $R$ is the radius of the Earth, and $h$ is the height above the Earth’s surface. Since the time period of the pendulum depends on the square root of the inverse of the acceleration due to gravity, any change in $g$ due to a change in height will affect the time period.

Given that the time period of the pendulum at a height $R$ above Earth's surface is $T_1$, and we're to find the time period $T_2$ at a height of $2R$, we can use the formula for acceleration due to gravity at different heights to express the relationship between $T_1$ and $T_2$.

For the initial case at height $R$:

$g_1 = g \left(\frac{R}{R + R}\right)^2 = g \left(\frac{R}{2R}\right)^2 = \frac{g}{4}$

For the new case at height $2R$:

$g_2 = g \left(\frac{R}{R + 2R}\right)^2 = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}$

The time period is proportional to the square root of the inverse of $g$, so:

$\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac{\frac{g}{9}}{\frac{g}{4}}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$

Therefore:

$T_1 = \frac{2}{3}T_2$

Rearranging this equation:

$3T_1 = 2T_2$

This corresponds to Option A.

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

Here energy depends on $A$ and $k$ and not on mass.

The initial momentum of the system (bullet + bob) is the momentum of the bullet because the bob is initially at rest. The momentum of the bullet is given by its mass times its velocity :

$ p_{\text{initial}} = m_{\text{bullet}} \times v_{\text{bullet}} $

After the collision, the bullet and the bob move together with a common velocity. Let's denote this common velocity as $ v' $. The final momentum $ p_{\text{final}} $ is the combined mass of the bullet and bob times the common velocity :

$ p_{\text{final}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v' $

According to the principle of conservation of linear momentum,

$ p_{\text{initial}} = p_{\text{final}} $

$ m_{\text{bullet}} \times v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v' $

Plugging in the values :

$ (10^{-2} \text{ kg}) \times (2 \times 10^2 \text{ m/s}) = (10^{-2} \text{ kg} + 1 \text{ kg}) \times v' $

Solving for $ v' $ :

$ v' = \frac{10^{-2} \times 2 \times 10^2}{10^{-2} + 1} = \frac{2}{1.01} \approx 1.98 \text{ m/s} $

After the collision, the system has some kinetic energy which will be completely converted to potential energy at the maximum height $ h $ that the bob reaches. Using the principle of conservation of energy :

$ \text{Kinetic Energy (KE)}_{\text{initial}} = \text{Potential Energy (PE)}_{\text{final}} $

$ \frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2 = (m_{\text{bullet}} + m_{\text{bob}})gh $

Isolating $ h $, we get :

$ h = \frac{\frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2}{(m_{\text{bullet}} + m_{\text{bob}})g} = \frac{v'^2}{2g} $

Substituting the values for $ v' $ and $ g $ :

$ h = \frac{(1.98)^2}{2 \times 10} = \frac{3.9204}{20} = 0.19602 \text{ m} $

Looking at the given options, the result most closely matches Option A :

$ \boxed{0.20 \text{ m}} $

Therefore, the height to which the bob rises before swinging back is approximately $ 0.20 \text{ m} $.

$\ell=10 \mathrm{~m}$,

Initial energy $=\mathrm{mg} \ell$

$\begin{aligned} & \text { So, } \frac{9}{10} \mathrm{mg} \ell=\frac{1}{2} \mathrm{mv}^2 \\ & \Rightarrow \frac{9}{10} \times 10 \times 10=\frac{1}{2} \mathrm{v}^2 \\ & \mathrm{v}^2=180 \\ & \mathrm{v}=\sqrt{180}=6 \sqrt{5} \mathrm{~m} / \mathrm{s} \end{aligned}$

In a simple pendulum experiment, time period $ T $ of a pendulum is given by,

$ T=2 \pi \sqrt{\frac{L}{g}} $

where, $ T= $ time period of pendulum, $ L= $ length of pendulum

$ g= $ acceleration due to gravity rearrange this formula to solve for $ g $,

$ \begin{array}{l} T^{2}=4 \pi^{2} \cdot \frac{L}{g} \\\\ g=4 \pi^{2} \cdot \frac{L}{T^{2}} \end{array} $

To find the percentage error in $ g $, differentiate the above equation and we get

$ \frac{\Delta g}{g} \times 100=\frac{\Delta L}{L} \times 100+2 \frac{\Delta T}{T} \times 100 $

Given, $ \frac{\Delta L}{L} \times 100=1 \% $ and $ \frac{\Delta T}{T} \times 100=2 \% $

So, $ \quad \frac{\Delta g}{g} \times 100=1 \%+2 \times 2 \% $

$ \begin{array}{l} \frac{\Delta g}{g} \times 100=(1+4) \% \\\\ \frac{\Delta g}{g} \times 100=5 \% \end{array} $

For a mass $ m $ oscillating with a spring of constant $ k $ and length $ l $, the time period $ T $ is given by,

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

When spring of length $ l $ is cut into three equal parts, each part has a length $ \frac{l}{3} $. The new spring constant, $ k^{\prime} $ is

$ k^{\prime}=3 k $

Equivalent spring constant for parallel combination

$ \begin{array}{l} \qquad \begin{array}{l} k_{\mathrm{eq}}=k^{\prime}+k^{\prime}+k^{\prime}=3 k^{\prime} \\\\ =3 \times 3 k=9 k \end{array} \end{array} $

Now, we load parallel combination with a mass of 4 m . The new time period using equivalent spring constant, is

$ T^{\prime}=2 \pi \sqrt{\frac{4 m}{k_{\mathrm{eq}}}}=2 \pi \sqrt{\frac{4 m}{9 k}}=\frac{4 \pi}{3} \sqrt{\frac{m}{k}} $

From Eq. (i),

$ T^{\prime}=\frac{2}{3} T $

The frequency of oscillation $ f^{\prime} $ is the reciprocal of the time period $ T^{\prime} $. Therefore,

$ \begin{array}{l} f^{\prime}=\frac{1}{T^{\prime}} \\\\ f^{\prime}=\frac{1}{\frac{2}{3} T} \\\\ f^{\prime}=\frac{3}{2 T} \end{array} $

Given, $ h=20 \mathrm{~m} $

$ u=0, v=31.4 \mathrm{~m} / \mathrm{s} $

Time period of simple pendulum $ =2 \mathrm{~s} $

Time period of simple pendulum is gives as

$ T=2 \pi \sqrt{\frac{l}{a}} $

( $ l= $ length, $ a= $ acceleration)

Using equation of motion to find $ a $

$ \begin{array}{l} v^{2}=u^{2}+2 a s \\\\ (31.4)^{2}=0+2 \times 20 \times a \\\\ (3.14)^{2} \times 100=40 a \\\\ a=(3.14)^{2} \times \frac{100}{40} \\\\ =(3.14)^{2} \times 2.5 \end{array} $

Substitute the value of $ a $ and $ T $ in Eq. (i)

$ 2=2 \pi \sqrt{\frac{1}{(3.14)^{2} \times 2.5}} $

$ \begin{array}{l} 1=\frac{3.14}{3.14} \sqrt{\frac{l}{2.5}} \\\\ 1=\sqrt{\frac{l}{2.5}} \end{array} $

Squaring both sides

$ l=2.5 \mathrm{~cm} $

Given, mass of particle,

$ m=4 \mathrm{mg}=4 \times 10^{-6} \mathrm{~kg} $

angular frequency, $\omega=40 \mathrm{rad} \mathrm{s}^{-1}$

Potential energy of the particle,

$ V=a+b x^2 $

We know that $F=-\frac{d V}{d x}=-2 b x$

also

$ F=-k x=-2 b x $

$ \Rightarrow \quad k=2 b $

Also $\frac{k}{m}=\omega^2$

$ \begin{aligned} \Rightarrow \quad k & =\omega^2 m=2 b \\\\ \Rightarrow \quad b & =\frac{\omega^2 m}{2} \\\\ & =\frac{(40)^2 \times 4 \times 10^{-6}}{2} \\\\ & =3200 \times 10^{-6} \mathrm{~J} / \mathrm{m}^2 \end{aligned} $

The mechanical energy $(E)$ of a damped oscillator is proportional to the square of its amplitude ( $A$ ). When the amplitude becomes half of its initial valuc, the mechanical energy decrease to a quarter of its initial value. This is

because the mechanical energy is given by

$ E \propto A^2 $

So, if the initial energy is $E_0$ and the initial amplitude is $A_0$. When the amplitude becomes $\frac{A_0}{2}$.

$ E=E_0\left(\frac{\frac{A_0}{2}}{A_0}\right)^2=E_0\left(\frac{1}{2}\right)^2=\frac{E_0}{4} $

$\therefore$ Decrease in mechanical energy of oscillator

$ =E_0-\frac{E_0}{4}=\frac{3 E_0}{4} $

$\therefore \%$ decrease in mechanical energy

$ =\frac{\frac{3 E_0}{4}}{E_0} \times 100=75 \% $

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.

To determine the time periods of vertical oscillations in different configurations of springs, we examine each setup.

Configuration (a):

The time period, $ T_a $, is given by the formula:

$ T_{a} = 2\pi \sqrt{\frac{m}{k}} $

Configuration (b):

In this setup, two identical springs are arranged in series, effectively halving the spring constant. Thus, the time period, $ T_b $, is:

$ T_{b} = 2\pi \sqrt{\frac{m}{\frac{k}{2}}} = 2\pi \sqrt{\frac{2m}{k}} $

Configuration (c):

In this setup, two identical springs are arranged in parallel, effectively doubling the spring constant. Hence, the time period, $ T_c $, is:

$ T_{c} = 2\pi \sqrt{\frac{m}{2k}} $

Now, calculating the ratios of these time periods, we get:

$ T_{a} : T_{b} : T_{c} = 1 : \sqrt{2} : \frac{1}{\sqrt{2}} $

This ratio expresses the relative oscillation times for each configuration of the springs.