Oscillations

$ \begin{aligned} &\text { Kinetic energy, } K=\frac{1}{2} m v^2\\ &\begin{aligned} & =\frac{1}{2} m A^2 \omega^2 \cos ^2(\omega t+\phi) \\ & \therefore K \propto \cos ^2(\omega t+\phi) \end{aligned} \end{aligned} $

At equilibrium position

Total energy = K.E

$ \begin{aligned} & \frac{1}{2} m v^2=0.02 \\ & \frac{1}{2} \times 20 \times v^2 \times 10^{-3}=2 \times 10^{-2} \\ & v=\sqrt{2} \\ & v=1.41 \mathrm{~m} / \mathrm{s} \end{aligned} $

Time taken for 30 oscillations $=60 \mathrm{~s}$

Time period of simple pendulum = Time taken for 1 oscillation

$ \begin{aligned} & \Rightarrow T=\frac{60}{30}=2 \mathrm{~s} \\ & T=2 \pi \sqrt{\frac{I}{g}} \Rightarrow I=\frac{g T^2}{4 \pi^2}=\frac{9.8 \times 2 \times 2}{4 \times 9.8}=1 \mathrm{~m} \end{aligned} $

At any point of time, time period is given by

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

Here $m$ is decreasing, so time period $T$ will be decreasing

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

Hence as mass leaks, $\omega$ will increase

Now, at any instant

$m g=k x_0$

So, equilibrium length $x_0=\frac{m g}{k}$, where $m$ is decreasing

So, equilibrium length will decrease.

So, amplitude also go on decreasing.

Two identical point masses, $P$ and $Q$, are suspended from two separate massless springs with spring constants $k_1$ and $k_2$, respectively. These masses oscillate vertically, and it is given that their maximum speeds are the same. We need to determine the ratio of the amplitude $A_Q$ of mass $Q$ to the amplitude $A_P$ of mass $P$.

The maximum velocity $ V $ for an oscillating mass is given by the equation $ V = A \omega $, where $ A $ is the amplitude and $ \omega $ is the angular frequency.

Given that the maximum velocities of $P$ and $Q$ are the same:

$ v_P = v_Q $

This implies:

$ A_P \omega_P = A_Q \omega_Q $

From this, the ratio of the amplitudes can be expressed as:

$ \frac{A_Q}{A_P} = \frac{\omega_P}{\omega_Q} $

The angular frequency $ \omega $ of a mass-spring system is given by:

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

Thus, for the two masses:

$ \omega_P = \sqrt{\frac{k_1}{m}} \quad \text{and} \quad \omega_Q = \sqrt{\frac{k_2}{m}} $

Substituting these into the ratio equation, we get:

$ \frac{A_Q}{A_P} = \frac{\sqrt{\frac{k_1}{m}}}{\sqrt{\frac{k_2}{m}}} = \sqrt{\frac{k_1}{k_2}} $

Hence, the ratio of the amplitude of mass $Q$ to the amplitude of mass $P$ is:

$ \sqrt{\frac{k_1}{k_2}} $

In simple harmonic motion (SHM), the total mechanical energy of the system is conserved and is a combination of kinetic energy (KE) and potential energy (PE). The total energy (E) of the particle can be expressed as:

$E = \frac{1}{2}kA^2,$

where $k$ is the spring constant, and $A$ is the amplitude of motion.

At a displacement $x$, the potential energy (PE) and kinetic energy (KE) of the particle are given by:

$\text{PE} = \frac{1}{2}kx^2$

$\text{KE} = E - \text{PE} = \frac{1}{2}kA^2 - \frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2)$

We need to find the displacement where the potential and kinetic energies are equal. This implies:

$\text{PE} = \text{KE}$

Thus, we get:

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

By simplifying, we get:

$x^2 = A^2 - x^2$

Adding $x^2$ to both sides:

$2x^2 = A^2$

Now, solving for $x$:

$x = \frac{A}{\sqrt{2}}$

Therefore, the displacement at which the potential and kinetic energies are equal is:

Option C:

$\frac{A}{\sqrt{2}}$

$\begin{aligned} & \vec{r}=(\hat{i}+2 \hat{j}) A \cos \omega t \\ & x=A \cos \omega t \\ & y=2 A \cos \omega t \\ & y=2 x \end{aligned}$

The path is straight line.

The motion is SHM and periodic as

$\begin{aligned} & \frac{d r}{d t}=-(\hat{i}+2 \hat{j}) \omega A \sin \omega t \\ & \frac{d^2 r}{d t^2}=-(\hat{i}+2 \hat{j}) \omega^2 A \cos \omega t \\ & \vec{a}=-\omega^2 \vec{r} \end{aligned}$

In the equation for simple harmonic motion (SHM), $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m},$ the general form $x = A \sin (\omega t + \phi)$ can be used to identify the parameters of SHM, where:

• A is the amplitude.
• ω (omega) is the angular frequency.
• φ (phi) is the phase constant.
• t is the time.

Comparing the given equation with the standard form:

• The amplitude A is 5 m, as that is the coefficient of sine in the equation.
• The angular frequency ω is $\pi$ rad/s.

The angular frequency $\omega$ is related to the time period $T$ of the motion through the formula:

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

Given that $\omega = \pi$, we can substitute and solve for $T$:

$\begin{aligned} \pi = \frac{2\pi}{T} & \implies T = \frac{2\pi}{\pi} = 2 \text{ seconds}. \end{aligned}$

Hence, the amplitude of the motion is 5 m and the time period is 2 s. Thus, the correct answer is:

Option B: 5 m, 2 s

The period of oscillation, $T$, of a simple pendulum is determined by the formula:

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

where:

• $L$ is the length of the pendulum
• $g$ is the acceleration due to gravity

The mass of the bob does not factor into the equation for the period.

Let's first denote the original length of the pendulum as $L$ and the original period of oscillation as $T_1$. Hence,

$ T_1 = 2\pi \sqrt{\frac{L}{g}} $

When the length of the pendulum is halved, the new length $L'$ would be $\frac{L}{2}$. Thus, the new period $T_2$ can be calculated as:

$ T_2 = 2\pi \sqrt{\frac{\frac{L}{2}}{g}} = 2\pi \sqrt{\frac{L}{2g}} = 2\pi \left(\frac{1}{\sqrt{2}}\right) \sqrt{\frac{L}{g}} = \frac{1}{\sqrt{2}} \cdot 2\pi \sqrt{\frac{L}{g}} = \frac{T_1}{\sqrt{2}} $

We are given that the new period $T_2$ is $\frac{x}{2} T_1$. Therefore, we can set up the equation:

$ \frac{T_1}{\sqrt{2}} = \frac{x}{2} T_1 $

To find the value of $x$, we solve for $x$:

$ \frac{1}{\sqrt{2}} = \frac{x}{2} $

Multiplying both sides by 2:

$ \frac{2}{\sqrt{2}} = x $

Simplify to:

$ x = \sqrt{2} $

Hence, the correct answer is:

Option B: $\sqrt{2}$

$ \begin{aligned} & \mathrm{T}_{\text {air }}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}=\sqrt{3} ~\mathrm{sec} \\ & \ln \text { Liquid } \rightarrow g_{\text {net }}=g\left(1-\frac{\rho}{\sigma}\right) \\ & \rho=\text { density of liquid } \\ & \sigma=\text { density of material of bob } \\ & \text { so } \mathrm{T}_{\text {Liq }}=2 \pi \sqrt{\left(\frac{\ell}{g_{\text {net }}}\right)}=2 \pi \sqrt{\frac{\ell}{g\left(1-\frac{\rho}{\sigma}\right)}} \\ & \mathrm{T}_{\text {Liq }}=\frac{\mathrm{T}_{\text {air }}}{\sqrt{1-\frac{\rho}{\sigma}}}=\frac{\sqrt{3}}{\sqrt{1-\frac{1}{4}}}=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=2 ~\mathrm{sec} \end{aligned} $

$ \begin{aligned} & \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}) \\ & \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}=\mathrm{A} \omega \cos (\omega \mathrm{t}) \\ & \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{a}=-\omega^{2} \mathrm{~A} \sin (\omega \mathrm{t}) \\ & \mathrm{a}=-\left(\frac{2 \pi}{8}\right)^{2} \times 1 \sin \left(\frac{2 \pi}{8} \times 2\right) \\ & \Rightarrow \mathrm{a}=-\frac{\pi^{2}}{16} \times \sin \left(\frac{\pi}{2}\right) \\ & \therefore \mathrm{a}=\frac{-\pi^{2}}{16} \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $

(a)

Amplitude of oscillation is continuously decreasing. It means bob of pendulum oscillate with air friction.

(b)

$F = - kx$ (restoring force of a spring)

(c)

Amplitude of oscillation is remains same. It means bob of pendulum is oscillating under negligible air resistance.

(d)

$P.E. = {1 \over 2}k{x^2},\,K.E. = {1 \over 2}m{\omega ^2}{A^2} - {1 \over 2}k{x^2}$

$T.E. = {1 \over 2}m{\omega ^2}{A^2} = $ constant

Here,

$\sin \omega t$ and $\sin \left( {\omega t + {\pi \over 4}} \right)$ clearly suggest a periodic sinusoidal function (SHM)

And $\sin \omega t + \cos \omega t = \sqrt 2 \sin \left( {\omega t + {\pi \over 4}} \right)$ is also a periodic sinusoidal function (SHM)

While, $y = {e^{ - \omega t}}$ is an exponentially decreasing function and thus does not have periodicity.

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

Let n₁ and n₂ be integer.

${n_1}{T_1} = {n_2}{T_2}$

$2\pi {n_1}\sqrt {{{1.21} \over g}} = 2\pi {n_2}\sqrt {{{1.00} \over g}} $

$ \Rightarrow {{{n_2}} \over {{n_1}}} = {{11} \over {10}}$

$\therefore$ After completion of 11^th oscillation of shorter pendulum, it will be in phase with longer pendulum.

At equilibrium,

$m g=A l \rho g \Rightarrow m=A \rho l$

where, $l$ length of part immersed in liquid.

When it is given in downward displacement $y$, restoring force (upward direction) on block is

$\begin{aligned} F & =-[A(l+y) \rho g-m g] \\\\ & =-[A(l+y) \rho g-A l \rho g] \\\\ & =-A \rho g y \end{aligned}$

i.e. $F \propto-y$ or $a \propto-\gamma$, so it execute SHM (inertia factor). Mass of block $=m$ Spring factor $=A \rho g$

$\begin{aligned} \text { Time period } & =2 \pi \sqrt{\frac{\text { Inertia factor }}{\text { Spring factor }}} \\\\ T & =2 \pi \sqrt{\frac{m}{A \rho g}} \\\\ \Rightarrow \quad T^2 & \propto \frac{m}{A \rho} \end{aligned}$

Time period of simple pendulum when it is falling under acceleration of ‘a’.

$T=2 \pi \sqrt{\frac{l}{g-a}}$ ..... (i)

In the case of freely falling, $a=g$

$\therefore$ From Eq. (i),

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

Given, spring constant $=k$

From law of conservation of energy

$m g h=\frac{1}{2} k x_0^2-m g x_0$

Where, $x_0$ is maximum elongation in spring

$\begin{array}{ll} \Rightarrow & \frac{1}{2} k x_0^2-m g x_0-m g h=0 \\ \Rightarrow & x_0^2-\frac{2 m g}{k} x_0-\frac{2 m g}{k} h=0 \\ & x_0=\frac{\frac{2 m g}{k} \pm \sqrt{\left(\frac{2 m g}{k}\right)^2+4 \times \frac{2 m g}{k} h}}{2} \end{array}$

Amplitude $=$ elongation in spring for lowest extreme position $-$ elongation in spring for equilibrium position

$=x_0-x_1=\frac{m g}{k} \sqrt{1+\frac{2 h k}{m g}}$

$\begin{gathered} \text {} k=\frac{f}{l} \Rightarrow k \propto \frac{1}{l} \\ \Rightarrow \frac{k_1}{k_2}=\frac{l_2}{l_1}=\frac{2}{1} \\ k_1=2 k, k_2=k \end{gathered}$

In series, $\frac{1}{k^{\prime}}=\frac{1}{k_1}+\frac{1}{k_2}=\frac{1}{2 k}+\frac{1}{k}=\frac{3}{2 k}$

$\therefore \quad k^{\prime}=2 / 3 k$