Simple Harmonic Motion

Since, $F+0.04 \pi^2 y=0$

$ \Rightarrow \quad F=-0.04 \pi^2 y $

For SHM, $F=-K y$

$ \begin{aligned} & \therefore & & K=0.04 \pi^2 \\ & \therefore & \omega & =\sqrt{\frac{K}{m}}=\sqrt{\frac{0.04 \pi^2}{0.09}}=\frac{\sqrt{4 \pi^2}}{9} \\ & \Rightarrow & \omega & =\frac{2 \pi}{3} \mathrm{rad} / \mathrm{s} \\ & \therefore & v_{\max } & =A \omega=\frac{6}{\pi} \times \frac{2 \pi}{3}=4 \mathrm{~m} / \mathrm{s} \end{aligned} $

For a damped oscillator

$ A=A_0 e^{-b t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \text { ( } b=\text { damping constant }) $

So, $A_1=A_0 e^{-b t_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

and $A_2=A_0 e^{-b t_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Multiplying Eqs. (ii) and (iii) we get,

$ \begin{aligned} & A_1 A_2=A_0^2 \cdot e^{-b\left(t_1+t_2\right)} \\ \Rightarrow \quad & \frac{A_1 A_2}{A_0}=A_0 e^{-b\left(t_1+t_2\right)} \end{aligned} $

Comparing with Eq. (i), amplitude at time $\left(t_1+t_2\right)$ is

$ \frac{A_1 A_2}{A_0} $

Step 1: Pendulum Frequency Formula

The time it takes for one swing (period) of a simple pendulum is given by $ T = 2\pi\sqrt{\frac{l}{g}} $, where $ l $ is length and $ g $ is gravity.

Frequency ($ f $) is how many swings it makes in one second, so $ f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{l}} $.

Step 2: Relationship Between Frequency and Length

From this, frequency is inversely related to the square root of length: $ f \propto \frac{1}{\sqrt{l}} $.

This means $ \frac{f_2}{f_1} = \sqrt{\frac{l_1}{l_2}} $, where $ l_1 $ and $ f_1 $ are initial length and frequency, and $ l_2 $ and $ f_2 $ are final.

Step 3: Assign the Values

The number of swings in a minute is frequency per minute. Original frequency = 72, new frequency = 90.

So, $ \frac{f_2}{f_1} = \frac{90}{72} = \frac{5}{4} $.

Now, $ \frac{5}{4} = \sqrt{\frac{l_1}{l_2}} $.

Step 4: Find the New Length

Square both sides: $ (\frac{5}{4})^2 = \frac{l_1}{l_2} $, so $ \frac{25}{16} = \frac{l_1}{l_2} $.

Rearrange to get $ l_2 = \frac{16}{25} l_1 $.

Step 5: Calculate the Percentage Decreased

Decrease in length = $ l_1 - l_2 $.

Percentage decrease = $ \frac{l_1 - l_2}{l_1} \times 100\% $.

Substituting values: $ \frac{l_1 - \frac{16}{25}l_1}{l_1} \times 100\% = \frac{9}{25} \times 100\% = 36\% $.

Understanding the Amplitude Change:

The formula for how the amplitude (A) of a damped oscillator changes over time is: $ A(t) = A_0 e^{-b t / 2 m} $ where A₀ is the starting amplitude, b is the damping constant, m is the mass, and t is time.

We are told the amplitude drops to half its original value in 10 seconds. So, $ A(t) = \frac{A_0}{2} $ when $ t = 10~\text{s} $.

Setting Up the Equation:

We set $ \frac{A_0}{2} = A_0 e^{-10b/2m} $

Divide both sides by A₀ to get: $ \frac{1}{2} = e^{-10b/2m} $

Take the natural logarithm of both sides: $ \ln(2) = \frac{10b}{2m} $

So, $ \frac{b}{m} = \frac{\ln(2)}{5} $  

Now, How Long for Energy to Halve?

Mechanical energy E is proportional to the square of amplitude: $ E \propto A^2 $. It also decays exponentially, but at twice the rate: $ E(t) = E_0 e^{-bt/m} $

We’re looking for time when $ E(t) = \frac{E_0}{2} $

So, $ \frac{E_0}{2} = E_0 e^{-bt/m} $

Divide by E₀: $ \frac{1}{2} = e^{-bt/m} $

Take the natural log: $ \ln(2) = \frac{bt}{m} $

Plug in the Value of $\\frac{b}{m}$:

From before, $ \frac{b}{m} = \frac{\ln(2)}{5} $, so substitute this into the energy equation:

$ \ln(2) = \left[ \frac{\ln(2)}{5} \right] t $

Divide both sides by $ \frac{\ln(2)}{5} $: $ t = 5~\text{seconds} $

Final Answer: It takes 5 seconds for the mechanical energy to become half of its original value.

The force in SHM is $F=-k x$

The maximum force is $F_{\text {max }}=-k A$

Given, $F=0.866 F_{\text {max }}$

$ \begin{aligned}So,\,\, -k x & =0.866(-k A) \\ x & =0.866 \mathrm{~A} \end{aligned} $

The velocity in SHM is $v=\omega \sqrt{\left(A^2-x^2\right)}$

$ \begin{aligned} & v=\omega \sqrt{A^2-(0.866 A)^2} \\ & v=\omega \sqrt{A^2-0.75 A^2}=\omega \sqrt{0.25 A^2} \\ & v=\omega(0.5 A) \end{aligned} $

The maximum velocity in SHM occurs at $x=0$, So $v_{\text {max }}=\omega A$

The ratio is $\frac{v}{v_{\text {max }}}=\frac{0.5 \omega A}{\omega A}$

$ \frac{v}{v_{\max }}=0.5=\frac{1}{2} $

The ratio of its velocity at that point and its maximum velocity is $\frac{1}{2}$.

Given:

$ \frac{U}{K} = \frac{4}{5} $

The amplitude $A = 6\ \mathrm{cm}$

Step 1: Write the energy formulas

The total energy is $E = \frac{1}{2} k A^2$

The potential energy is $U = \frac{1}{2} k x^2$

The kinetic energy is $K = \frac{1}{2} k (A^2 - x^2)$

Step 2: Set up the given ratio

We want the distance $x$ from the mean position so that $\frac{U}{K} = \frac{4}{5}$:

$ \frac{\frac{1}{2}k x^2}{\frac{1}{2}k (A^2 - x^2)} = \frac{4}{5} $

Step 3: Simplify the equation

The $\frac{1}{2}k$ cancels out: $ \frac{x^2}{A^2 - x^2} = \frac{4}{5} $

Step 4: Solve for $x^2$

Multiply both sides by $(A^2 - x^2)$: $ 5 x^2 = 4 (A^2 - x^2) $

Expand: $ 5 x^2 = 4 A^2 - 4 x^2 $

Bring $4x^2$ to the left: $ 5 x^2 + 4 x^2 = 4 A^2 $

Combine like terms: $ 9 x^2 = 4 A^2 $

Step 5: Find $x$

Divide both sides by 9: $ x^2 = \frac{4}{9} A^2 $

Take the square root: $ x = \frac{2}{3} A $

Since $A = 6\ \mathrm{cm}$:

$ x = \frac{2}{3} \times 6 = 4\ \mathrm{cm} $

The distance from the mean position is $4~\mathrm{cm}$.

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 \% $

Given

$ x=4(\cos \pi t+\sin \pi t) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

The standard equation for amplitude is

$ x=A(\sin \omega t+\phi) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

We have to correct Eq. (i) in format of Eq. (ii) to compare the amplitude.

$x=4[\cos \pi t+\sin \pi t] \times \frac{\sqrt{2}}{\sqrt{2}}$ [multiply and divide it by $\sqrt{2}$ ]

$ \begin{aligned} & x= 4 \sqrt{2}\left[\frac{1}{\sqrt{2}} \cos \pi t+\frac{1}{\sqrt{2}} \sin \pi t\right] \\ & x= 4 \sqrt{2}\left[\cos \pi t \sin \frac{\pi}{4}+\sin \pi t \cos \frac{\pi}{4}\right] \\ & {\left[\frac{1}{\sqrt{2}}=\sin \frac{\pi}{4} \text { and } \cos \frac{\pi}{4}\right] } \\ & x=4 \sqrt{2} \sin \left(\pi t-\frac{\pi}{4}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

because $[\sin (A+B)=[\sin B \cos A +\cos B \sin A] $

Compare Eqs. (iii) and (ii)

$ A=4 \sqrt{2} $

Hence, the amplitude is $4 \sqrt{2}$.

Given,

SHM displacement $x=2 \cos (t)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Compare the displacement with standard displacement of SHM

$ x=A \cos (k x+\omega t)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

Now comparing Eq. (i) and (ii)

$ A=2, K=0, \omega=1 $

where $\omega$ is angular frequency.

$\omega=\frac{2 \pi}{T}$ where $T$ is time period.

$ T=\frac{2 \pi}{\omega} \Rightarrow \frac{2 \pi}{1} s $

So, the time period of particle is $2 \pi$ seconds.

Given force, $F=6.4 \mathrm{~N}$

Elongation in spring $(\Delta x)=0.1 \mathrm{~m}$

time period, $T=\pi / 4$

Time period of oscillation is given by

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

Since we know, $F=K \Delta x$

$ \begin{array}{cc} & 6.4=K(0.1) \\ \Rightarrow & K=64 \mathrm{~N} / \mathrm{m} \\ \text { Thus, } & T=2 \pi \sqrt{\frac{m}{64}} \Rightarrow \frac{\pi}{4}=\frac{2 \pi \sqrt{m}}{8} \\ \Rightarrow & m=1 \mathrm{~kg} \end{array} $

The effective acceleration of a bob in water is given by,

$ g^{\prime}=g\left(1-\frac{\sigma}{\rho}\right) $

Where, $\sigma$ and $\rho$ are densities of water and bob respectively.

Since, the periods of oscillation of bob in air and water are given as

$ T=2 \pi \sqrt{\frac{l}{g}} \text { and } T=2 \pi \sqrt{\frac{l}{g^{\prime}}} $

$ \begin{aligned} & \text { We have, } \frac{T}{T^{\prime}}=\sqrt{\frac{g^{\prime}}{g}} \\ & \quad \frac{T}{T^{\prime}}=\sqrt{\frac{g\left(1-\frac{\sigma}{\rho}\right)}{g}} \\ & \Rightarrow \quad \frac{T}{T^{\prime}}=\sqrt{1-\frac{\sigma}{\rho}}=\sqrt{1-\frac{1}{\rho}} \quad(\because \sigma=1) \\ & \quad \frac{T}{T^{\prime}}=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{1}{\rho}} \\ & \Rightarrow \quad \frac{1}{2}=1-\frac{1}{\rho} \\ & \Rightarrow \quad \rho=2 \end{aligned} $

Initial Setup:

Spring constant: $k$

Mass: $m$

Initial time period: $1 \text{ s}$

New Setup:

Replaced mass: $4m$

For a spring-block system in simple harmonic motion (SHM), the time period $t$ is given by:

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

In the initial setup, we have:

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

Rearranging gives:

$ \frac{1}{4\pi^2} = \frac{m}{k} \quad \Rightarrow \quad k = \frac{m}{4\pi^2} $

Now, substituting this new mass into the time period formula:

$ t_2 = 2\pi \sqrt{\frac{4m}{k}} $

Replacing $k$ with the expression found:

$ t_2 = 2\pi \sqrt{\frac{4m}{\frac{m}{4\pi^2}}} = 2\pi \sqrt{\frac{4\times4\pi^2}{1}} = 2\pi \sqrt{16\pi^2} $

This simplifies to:

$ t_2 = 0.5 \text{ s} $

Thus, the clock now runs slow by 0.5 seconds for every second that passes.

Given,

Particle is executing SHM at $x_1=40 \mathrm{ml}$ from mean position.

$ \mathrm{KE}=\frac{1}{3} \mathrm{KE}_{\max } $

max potential energy of SHM$\begin{aligned} & \mathrm{KE}_{\max }=\mathrm{TE}=\frac{1}{2} K A^2 \\ & \mathrm{KE}=\frac{1}{3}\left(\mathrm{KE}_{\max }\right)=\frac{1}{3} \times \frac{1}{2} K A^2 \\ &=\frac{1}{3} \times \frac{1}{2} K A^2 \\ & \frac{1}{2} K\left(A^2-x^2\right)=\frac{1}{6} K A^2 \\ & A^2-x^2=\frac{1}{3} A^2 \\ & { At } x=4 \mathrm{~cm} \\ & A^2-16=\frac{1}{3} A^2 \\ & \therefore A=2 \sqrt{6} \mathrm{~cm}\end{aligned}

Total mechanical energy for a block in SHM is

$ E=\frac{1}{2} m \omega^2 a^2 $

Given, $E=100 \mathrm{~J}, a=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$

So, $\quad 100=\frac{1}{2} m \omega^2 \times\left(5 \times 10^{-2}\right)^2$

$ \begin{aligned} & \Rightarrow \quad \frac{1}{2} m \omega^2=\frac{100}{25 \times 10^{-4}} \\ & \Rightarrow \quad \frac{1}{2} m \omega^2=4 \times 10^4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, at $t=0$, displacement $x$ is given by

$ \begin{aligned} x & =5 \cos \left(\omega t+\frac{\pi}{4}\right) \\ \Rightarrow x(t=0) & =5 \cos \left(\frac{\pi}{4}\right)=\frac{5}{\sqrt{2}} \mathrm{~cm} \end{aligned} $

Now, potential energy at, $x=\frac{5}{\sqrt{2}} \times 10^{-2}$ m is

$ \begin{aligned} U & =\frac{1}{2} m \omega^2 \cdot x^2 \\ & =4 \times 10^4 \times\left(\frac{5}{\sqrt{2}} \times 10^{-2}\right)^2 \\ & =\frac{4 \times 25}{2}=50 \mathrm{~J} \end{aligned} $

Let displacement of particle in SHM is

$ x=A \cos (\omega t+\phi) $

Then, it's velocity is

$ v=\frac{d x}{d t}=-A \omega \sin (\omega t+\phi) $

Now, at $t=2 \mathrm{~s}, x=0$

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

where $T=16 \mathrm{~s}$

$ \begin{array}{llrl} & \text { So, } 0= & A \cos \left(\frac{2 \pi}{16} \times 2+\phi\right) \\ \Rightarrow & & \cos \left(\frac{\pi}{4}+\phi\right) & =0 \\ \Rightarrow & & \frac{\pi}{4}+\phi & =\frac{\pi}{2} \\ \Rightarrow & & \phi & =\frac{\pi}{4} \end{array} $

Also, velocity is $4 \mathrm{~m} / \mathrm{s}$ at $t=4 \mathrm{~s}$. So, $v=-A \omega(\sin (\omega t+\phi))$ gives,

$ \begin{array}{ll} \Rightarrow & 4=-A\left(\frac{2 \pi}{16}\right) \cdot \sin \left(\frac{2 \pi}{16} \times 4+\frac{\pi}{4}\right) \\ \Rightarrow & 4=-A \times \frac{2 \pi}{16} \times \sin \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \\ \Rightarrow & 4=-A \times \frac{2 \pi}{16} \times \cos \frac{\pi}{4} . \\ \Rightarrow & 4=\frac{-32 \sqrt{2}}{\pi} \end{array} $

or magnitude of amplitude is $\frac{32 \sqrt{2}}{\pi}$ units.

Given,

$ A(t)=A_0 e^{b t / 2 m} $

where, $b=70 \mathrm{~g} / \mathrm{s}, m=200 \mathrm{~g}$

Mechanical energy ( $E$ ) of the damped oscillation at any instant $t$ is given by

$ \begin{aligned} & & E & =E_0 e^{-b t / m} \\ & \text { Here, } & E & =\frac{E_0}{4}, t=T^{\prime} \\ & \therefore & \frac{E_0}{4} & =E_0 e^{-b T^{\prime} / m} \\ & \Rightarrow & 4 & =e^{b T^{\prime} / m} \\ & \Rightarrow & \log _e 4 & =\frac{b T^{\prime}}{m} \\ & \Rightarrow & 2 \log _e 2 & =\frac{b T^{\prime}}{m} \end{aligned} $

$ \, \begin{aligned}\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad T^{\prime} & =2 \times 2.3026 \log _{10} 2 \times \frac{m}{b} \\ & =2 \times 2.3026 \times 0.3010 \times \frac{200}{70} \\ & =3.96 \mathrm{~s} \simeq 4 \mathrm{~s} \end{aligned} $

Length of simple pendulum, *

$ l=1 \mathrm{~m} $

Mass of bob, $m=100 \mathrm{~g}=0.1 \mathrm{~kg}$

Radius of circular track, $R=100 \mathrm{~m}$

Speed, $v=10 \mathrm{~m} / \mathrm{s}$

The bob of the simple pendulum will experience acceleration due to gravity which provide centripetal acceleration in the circular motion of the car.

Centripetal acceleration, $a_c=v^2 / R$

where, $v$ is the uniform speed of the car and $R$ is the radius of track.

∴ Effective value of acceleration,

$ \begin{aligned} & \begin{aligned} g^{\prime} & =\sqrt{g^2+\left(a_c\right)^2}=\sqrt{g^2+\left(\frac{v^2}{R}\right)^2} \\ \Rightarrow \quad g^{\prime} & =\sqrt{g^2+\frac{v^4}{R^2}} \end{aligned} \\ & \therefore \text { Time period, } T =2 \pi \sqrt{\frac{l}{g^{\prime}}} \\ & =2 \pi \sqrt{\frac{l}{\sqrt{g^2+\frac{v^4}{R^2}}}} \\ & =2 \pi \sqrt{\frac{1}{\sqrt{10^2+\frac{10^4}{(100)^2}}}} \\ & =2 \pi \sqrt{\frac{1}{\sqrt{10^2+1}}} \end{aligned} $

$ \begin{array}{rlrl} \Rightarrow & T =\frac{2 \pi}{\sqrt{\frac{1}{\sqrt{101}}}}=\frac{2 \pi}{(101)^{1 / 4}} \\ \,\,\,\,\,\,& =\frac{2 \pi}{\alpha^{1 / 4}} \\ & \therefore \alpha =101 \end{array} $

The given situation is shown below.

$ \begin{aligned} &\text { Effective gravitational acceleration }\\ &\begin{aligned} & =\left(g-\frac{q E}{m}\right) \\ \therefore \quad T & =2 \pi \sqrt{\frac{l}{g-\frac{q E}{m}}} \end{aligned} \end{aligned} $

$ \text { The given situation is shown as } $

Let the angle be $\omega t$ when the KE is $75 \%$ of total energy.

Also, the moment of inertia of the bob of pendulum about the point of suspension

$ =I=m l^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, velocity at that instant

$ =l \omega \cos \omega t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, according to the question,

$ \frac{1}{2} m v^2=\frac{75}{100} \times \frac{1}{2} I \omega^2 $

where, $\left(\frac{1}{2} I \omega^2\right)$ is the rotational KE of the particle,

$ \begin{aligned} & \frac{1}{2} m v^2=\frac{75}{100} \times \frac{1}{2} \times m l^2 \times \omega^2 \text { [from Eq. (i)] } \\ & \Rightarrow \frac{1}{2} l^2 m \omega^2 \cos ^2 \omega t=\frac{3}{4} \times \frac{1}{2} \times m l^2 \omega^2 \end{aligned} $

[from Eq. (ii)]

$ \begin{aligned} & \Rightarrow \quad \cos ^2 \omega t=3 / 4 \\ & \Rightarrow \quad \cos \omega t= \pm \frac{\sqrt{3}}{2} \Rightarrow \omega t=\frac{\pi}{6} \end{aligned} $

(keeping only positive solution)

$ t=\frac{\pi}{6 \omega}=\frac{\pi T}{6 \times 2 \pi} $

( $\therefore$ Time period, $T=\frac{2 \pi}{\omega}$ )

$ \begin{array}{ll} \Rightarrow & t=\frac{\pi \times 4}{12 \pi} \,\,(given, T=4 \mathrm{~s} )\\ \Rightarrow & t=\frac{1}{3} \mathrm{~s} \end{array} $

The frequency of oscillation

$ \begin{equation*} \omega=\sqrt{\frac{K}{m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

The maximum acceleration in SHM,

$ \alpha_{\max }=\omega^2 \chi $

where, $x$ is the maximum extension in the spring.

To maintain contact with spring, the maximum SHM acceleration should be equal to acceleration due to gravity.

$ \text { } \quad \begin{align*} i.e., \omega^2 x & \left.=g \Rightarrow x=\frac{m g}{K} \text { [from Eq. (i)] }\right. \\ & =\frac{10 \times 10}{400}=\frac{1}{4}[\therefore K=400 \mathrm{~N} / \mathrm{m}] \\ & =0.25 \mathrm{~m} \text { or } 25 \mathrm{~cm} \end{align*} $

Given, amplitude,

$ A=0.2 \mathrm{~cm}=2 \times 10^{-3} \mathrm{~m} $

Velocity at mean position in SHM

$ v_{\max }=5 \mathrm{~m} / \mathrm{s} $

We know that,

$ \begin{aligned} v_{\max } & =A \omega \Rightarrow \omega=\frac{v_{\max }}{A} \\ & =\frac{5}{2 \times 10^{-3}}=25 \times 10^3 \\ & =2500 \mathrm{rad} \mathrm{~s}^{-1} \end{aligned} $

Displacement equation of simple harmonic motion is given as

$ \begin{aligned} x= & 6 \cos \left(2 \pi t+\frac{\pi}{3}\right) \\ \therefore \text { Velocity, } v & =\frac{d x}{d t}=\frac{d}{d t}\left[6 \cos \left(2 \pi t+\frac{\pi}{3}\right)\right] \\ & =-6 \sin \left(2 \pi t+\frac{\pi}{3}\right) \times 2 \pi \\ v & =-12 \pi \sin \left(2 \pi t+\frac{\pi}{3}\right) \end{aligned} $

$ \begin{aligned} & \therefore \text { Acceleration, } a=\frac{d v}{d t} \\ & =\frac{d}{d t}\left[-12 \pi \sin \left(2 \pi t+\frac{\pi}{3}\right)\right] \\ & =-12 \pi \cos \left(2 \pi t+\frac{\pi}{3}\right)(2 \pi) \end{aligned} $

$ \begin{aligned} & a=-24 \pi^2 \cos \left(2 \pi t+\frac{\pi}{3}\right) \\ & \text { At, } t=1 \mathrm{~s} \\ & a=-24 \pi^2 \cos \left(2 \pi+\frac{\pi}{3}\right) \\ & =-24 \pi^2 \cos \frac{\pi}{3}=-24 \pi^2 \times \frac{1}{2}=-12 \pi^2 \\ & \therefore|a|=12 \pi^2 \mathrm{~ms}^{-2} \end{aligned} $

Displacement of particle is

$ x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right) $

Velocity, $v=\frac{d x}{d t}$

$ \Rightarrow v=-x_0 \omega \sin \left(\omega t-\frac{\pi}{4}\right) $

Acceleration, $a=\frac{d v}{d t}$

$ \begin{aligned} & =-x_0 \omega^2 \cos \left(\omega t-\frac{\pi}{4}\right) \\ & =x_0 \omega^2 \cdot \cos \left(\pi+\left(\omega t-\frac{\pi}{4}\right)\right) \\ & =x_0 \omega^2 \cdot \cos \left(\omega t+\frac{3 \pi}{4}\right) \\ & a=x_0 \omega^2 \cos \left(\omega t-\left(-\frac{3 \pi}{4}\right)\right) \end{aligned} $

Comparing with $a=A \cos (\omega t-\delta)$, we have

$ A=x_0 \omega^2, \delta=-\frac{3 \pi}{4} $

Average acceleration of particle in SHM between extreme and equilibrium position is

$ A_{\mathrm{avg}}=\frac{\omega^2 a}{\pi / 2}=\omega^2 a \times \frac{2}{\pi}\left(\mathrm{~ms}^{-2}\right) $

where, $\omega^2 a=A_{\text {max }}$

So, $\frac{A_{\text {avg }}}{A_{\text {max }}}=\frac{2}{\pi}$