Oscillations

\begin{equation}
\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}} \\
& \mathrm{~g}=\frac{4 \pi^2 \ell}{\mathrm{~T}^2} \\
& \text { Percentage error, } \frac{\Delta \mathrm{g}}{\mathrm{~g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{~T}}{\mathrm{~T}} \\
& =\frac{0.2}{20}+2\left(\frac{1}{40}\right)=\frac{1.2}{20}
\end{aligned}
\end{equation}

percentage change = $\frac{1.2}{20}\times 100=6\%$

At altitude \(h=2R\) above Earth’s surface, the distance from Earth’s center is \(r=R+h=3R\).
Acceleration due to gravity there is
\[
g' = g\left(\frac{R}{r}\right)^2 = g\left(\frac{R}{3R}\right)^2 = \frac{g}{9}.
\]

For a seconds pendulum \(T=2\,\text{s}\) and
\[
T=2\pi\sqrt{\frac{L}{g'}} \;\Rightarrow\; L=\frac{g' T^2}{4\pi^2}.
\]
With \(g=\pi^2\,\text{m s}^{-2}\),
\[
L=\frac{\left(\dfrac{g}{9}\right)\, (2)^2}{4\pi^2}
=\frac{g}{9}\cdot\frac{4}{4\pi^2}
=\frac{g}{9\pi^2}
=\frac{\pi^2}{9\pi^2}
=\frac{1}{9}\ \text{m}.
\]

\[
\boxed{\dfrac{1}{9}\ \text{m}}
\]

At rest (v = 0):

$\cos (50t+\pi /3)=0\Rightarrow 50t+\pi /3=\frac{\pi }{2}$$t_1=\frac{\pi /2-\pi /3}{50}=\frac{\pi }{300}$

Acceleration $\left(a=-{\omega }^{2}x\right)$ :

$a=-2500a\sin (50t+\pi /3)$

For zero acceleration:

$\sin (50t+\pi /3)=0\Rightarrow 50t+\pi /3=\pi$

$t_2=\frac{\pi -\pi /3}{50}=\frac{\pi }{75}$

The period T is 2 seconds, so the number of complete cycles in 12.5 seconds is:

$n=\frac{12.5}{2}=6.25\text{ cycles. }$

In one full cycle, the particle moves a distance of 4 A (since it goes from +A to -A and back to +A).
So, in 6.25 cycles, the total distance covered is:

$D=6.25\times 4A=6.25\times 4\times 1=25cm$

Since after 6 full cycles, the particle returns to its starting position, it only moves A in the last incomplete cycle.
Therefore, the displacement is d =1 cm.

$\frac{D}{d}=\frac{25}{1}=25$

$$\begin{gathered} T=2\pi \sqrt{\frac{\ell }{g}} \\ g=\frac{g_0{R}^2}{(R+h{)}^2} \end{gathered}$$

As h increases, g decreases, and T increases.

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

$g_2=\frac{GM}{{\left(R+2R\right)}^2}$

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

$\Rightarrow 3T_1=2T_2$

For a simple harmonic motion (SHM), the maximum speed $V_{max}$​ of a particle is given by:

$V_{max}=A\omega$

where:

• $A$ is the amplitude of oscillation.

• $\omega$ is the angular frequency, given by:

  $$\omega = \sqrt{\frac{k}{m}}$$​

Thus, the maximum speed becomes:

$V_{max}=A\sqrt{\frac{k}{m}}$​

Let the two particles have spring constants $K_1$​ and $K_2$ but the same mass $m$ and same amplitude $A$.

For the first particle:

$$v_{\max1} = A \sqrt{\frac{K_1}{m}}$$​

For the second particle:

$$v_{\max2} = A \sqrt{\frac{K_2}{m}}$$

Taking the ratio:

$$\frac{v_{\max1}}{v_{\max2}} = \frac{A \sqrt{\frac{K_1}{m}}}{A \sqrt{\frac{K_2}{m}}}$$​​

Canceling $A$A and simplifying:

$$\frac{v_{\max1}}{v_{\max2}} = \sqrt{\frac{K_1}{K_2}}$$

$$\boxed{A} \quad \sqrt{\frac{K_1}{K_2}}$$

The equation for the time period T of the oscillations is given as:

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

Using

$$\begin{gathered} a^{2}xrg=m{a}_{\text{net }} \\ \frac{L^2\rho g}{m}x=a_{\text{net }} \\ m=10g,L=10cm,\rho =1000kg/m^3 \end{gathered}$$

and putting values in equation

$$\begin{gathered} T=2\pi \times {10}^{-2}S \\ y=2 \end{gathered}$$

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

$g_2=\frac{GM}{{\left(R+2R\right)}^2}$

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

$\Rightarrow 3T_1=2T_2$

Given Data

Mass $m=200\text{ g}=0.2\text{ kg}$

Equilibrium extension $x_0=2\text{ mm}=2\times {10}^{-3}\text{ m}$

Amplitude (further pull) $A=2\text{ mm}=2\times {10}^{-3}\text{ m}$

Find the spring constant (\(k\))

At the initial equilibrium position, the spring force balances the weight:

Calculate the frequency (\(f\))

Using the standard formula for the frequency of a spring-mass system:

Calculate the maximum energy in the spring

Key Concept: The maximum energy stored in the spring occurs at the lowest point of oscillation, where the total extension is the sum of the equilibrium stretch and the amplitude ($x_{max}=x_0+A$).

Now, find the maximum elastic potential energy:

Correct Option: A

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

$g_2=\frac{GM}{{\left(R+2R\right)}^2}$

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

$\Rightarrow 3T_1=2T_2$

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

$g_2=\frac{GM}{{\left(R+2R\right)}^2}$

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

$\Rightarrow 3T_1=2T_2$

Since the particle is moving from the extreme position (\(x = A\)), we use the cosine equation for simple harmonic motion:

Substitute the target position $x=\frac{A}{\sqrt{2}}$:

This gives the phase angle:

Substitute angular frequency $\omega =\frac{2\pi }{T}$:

Solving for \(t\):

Given the time period \(T = 5\) sec:

The acceleration due to gravity:

$g=\frac{GM}{R^2}$

gravity on the new planet:

$g^{\text{'}}=\frac{G\left(4M\right)}{(2R{)}^2}=\frac{4GM}{4R^2}=\frac{GM}{R^2}=g$

Assertion (a) is true.
The mass of the pendulum does not affect the time period. While it is true that the mass remains unchanged, it is not the reason why the time period remains the same.

$$\begin{gathered} T=2\pi \sqrt{\frac{\ell }{g}} \\ g=\frac{g_0{R}^2}{(R+h{)}^2} \end{gathered}$$

As h increases, g decreases, and T increases.

For a spring,

$k\ell =\text{ constant }$

If the spring length is halved:

$k_2=2k_1$

Frequency:

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

Hence,

$\frac{v_2}{v_1}=\sqrt{\frac{k_2}{k_1}}=\sqrt{2}$

For a simple harmonic motion (SHM), the maximum speed $V_{max}$​ of a particle is given by:

$V_{max}=A\omega$

where:

• $A$ is the amplitude of oscillation.

• $\omega$ is the angular frequency, given by:

  $$\omega = \sqrt{\frac{k}{m}}$$​

Thus, the maximum speed becomes:

$V_{max}=A\sqrt{\frac{k}{m}}$​

Let the two particles have spring constants $K_1$​ and $K_2$ but the same mass $m$ and same amplitude $A$.

For the first particle:

$$v_{\max1} = A \sqrt{\frac{K_1}{m}}$$​

For the second particle:

$$v_{\max2} = A \sqrt{\frac{K_2}{m}}$$

Taking the ratio:

$$\frac{v_{\max1}}{v_{\max2}} = \frac{A \sqrt{\frac{K_1}{m}}}{A \sqrt{\frac{K_2}{m}}}$$​​

Canceling $A$A and simplifying:

$$\frac{v_{\max1}}{v_{\max2}} = \sqrt{\frac{K_1}{K_2}}$$

$$\boxed{A} \quad \sqrt{\frac{K_1}{K_2}}$$

1. Key concepts:

   • The total distance $D$ is the sum of all oscillations' displacements (absolute values of motion).
   • Displacement $d$d is the shortest distance between the initial and final positions.
   • The time period $T = 2 \, \text{s}$.

2. Oscillation in 12.5 seconds:

   • Number of complete oscillations: $$n = \frac{\text{Total time}}{\text{Time period}} = \frac{12.5}{2} = 6.25$$
   • This implies 6 full oscillations and 1/4 of another oscillation.

3. Displacement ($d$):

   • After 6 full oscillations, the block returns to its starting position.
   • During the remaining $0.25 \times T = 0.5 \, \text{s}$, the block moves from the mean position to the maximum displacement (amplitude $A$).
   • Hence, $d = A$

4. Total distance ($D$):

   • In one complete oscillation, the block covers a distance of $4A$(2A forward and 2A backward).
   • For 6 complete oscillations: $$D_{\text{6 oscillations}} = 6 \cdot 4A = 24A$$
   • In the last quarter oscillation: $$D_{\text{last quarter}} = A$$
   • Total distance: $$D = D_{\text{6 oscillations}} + D_{\text{last quarter}} = 24A + A = 25A$$

5. Ratio $\frac{D}{d}$​:

   $$\frac{D}{d} = \frac{25A}{A} = 25$$

Answer: $\frac{D}{d} = 25$

The period T is 2 seconds, so the number of complete cycles in 12.5 seconds is:

$n=\frac{12.5}{2}=6.25\text{ cycles. }$

In one full cycle, the particle moves a distance of 4 A (since it goes from +A to -A and back to +A).
So, in 6.25 cycles, the total distance covered is:

$D=6.25\times 4A=6.25\times 4\times 1=25cm$

Since after 6 full cycles, the particle returns to its starting position, it only moves A in the last incomplete cycle.
Therefore, the displacement is d =1 cm.

$\frac{D}{d}=\frac{25}{1}=25$

$\text{ (*) (A) }{\sin }^2\omega t=\frac{1}{2}(1-\cos 2\omega t)$
This is periodic with period
$T=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }(\mathrm{it} \mathrm{is} \mathrm{a} \mathrm{SHM})$
$\text{ (B) }{\sin }^3\left(2\omega t\right)=\frac{3}{4}\sin \left(2\omega t\right)-\frac{1}{4}\sin \left(6\omega t\right)$
The fundamental angular frequency is $2\omega$
It is periodic with common period $T=\frac{\pi }{\omega }$ but not SHM.
$\text{ (C) }\sin \left(\omega t\right)+\cos \left(\pi \omega t\right)$
In general, these two terms do not have
a common period, so it is non-periodic.
$\text{ (D) }\cos \omega t+\cos 2\omega t$
$\text{ Common period: }T=\frac{2\pi }{\omega }$
It is periodic, but not SHM.
Hence,
A → III, B → I, C → IV, D → II
No option in correct.

\begin{equation}
\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}} \\
& \mathrm{~g}=\frac{4 \pi^2 \ell}{\mathrm{~T}^2} \\
& \text { Percentage error, } \frac{\Delta \mathrm{g}}{\mathrm{~g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{~T}}{\mathrm{~T}} \\
& =\frac{0.2}{20}+2\left(\frac{1}{40}\right)=\frac{1.2}{20}
\end{aligned}
\end{equation}

percentage change = $\frac{1.2}{20}\times 100=6\%$

The period T is 2 seconds, so the number of complete cycles in 12.5 seconds is:

$n=\frac{12.5}{2}=6.25\text{ cycles. }$

In one full cycle, the particle moves a distance of 4 A (since it goes from +A to -A and back to +A).
So, in 6.25 cycles, the total distance covered is:

$D=6.25\times 4A=6.25\times 4\times 1=25cm$

Since after 6 full cycles, the particle returns to its starting position, it only moves A in the last incomplete cycle.
Therefore, the displacement is d =1 cm.

$\frac{D}{d}=\frac{25}{1}=25$

(B)