Oscillations Practice Questions

The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$, which implies $T \propto \sqrt{L}$.
Given the time duration $t$ is constant, the number of oscillations $N$ relates to the time period as $T = \frac{t}{N}$, implying $N \propto \frac{1}{\sqrt{L}}$, or $L \propto \frac{1}{N^2}$.
For the two cases, we have $\frac{L_1}{L_2} = \left( \frac{N_2}{N_1} \right)^2$.
Substituting the given values, $L_1 = 30\text{ cm}$, $N_1 = 20$, and $N_2 = 40$:
$\frac{30}{L_2} = \left( \frac{40}{20} \right)^2 = 2^2 = 4$.
Solving for $L_2$ gives $L_2 = \frac{30}{4} = 7.5\text{ cm}$.

For a two-body system connected by a spring, the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{\mu}}$, where $\mu$ is the reduced mass.
Given values are $m_1 = 1 \text{ kg}$, $m_2 = 0.2 \text{ kg}$, and $k = 150 \text{ N/m}$.
First, calculate the reduced mass: $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{1 \times 0.2}{1 + 0.2} = \frac{0.2}{1.2} = \frac{1}{6} \text{ kg}$.
Now, substitute $\mu$ and $k$ into the angular frequency formula: $\omega = \sqrt{\frac{150}{1/6}}$.
Simplifying this expression yields $\omega = \sqrt{150 \times 6} = \sqrt{900} = 30 \text{ rad/s}$.

When the cylindrical block of mass $M$ is depressed by an additional displacement $x$, the increased submerged volume $A x$ generates an additional upward buoyant force $F = \rho A x g$. This force acts as a restoring force, so $F = -(\rho A g)x$. Applying Newton's second law, $M \frac{d^2x}{dt^2} = -(\rho A g)x$, which describes Simple Harmonic Motion (SHM) of the form $\frac{d^2x}{dt^2} + \omega^2 x = 0$. Comparing these equations, the angular frequency is $\omega = \sqrt{\frac{\rho A g}{M}}$. Consequently, the period of oscillation $T = \frac{2\pi}{\omega}$ is given by $T = 2\pi \sqrt{\frac{M}{\rho A g}}$.

The time period $T$ of a simple pendulum of length $L$ is given by the formula:
$T = 2\pi \sqrt{\frac{L}{g}}$
Squaring both sides of the equation yields:
$T^2 = \frac{4\pi^2 L}{g}$
Taking the reciprocal of both sides, we get:
$\frac{1}{T^2} = \frac{g}{4\pi^2} \cdot \frac{1}{L}$
This equation is of the form $y = \frac{k}{x}$ (where $k = \frac{g}{4\pi^2}$ is a constant), which describes an inverse proportionality between $\frac{1}{T^2}$ and $L$. The graph of this relationship is a rectangular hyperbola, which corresponds to the first plot.

The angular frequency of the simple harmonic oscillator is given as $\omega = 176 \text{ rad/s}$.
The relationship between angular frequency $\omega$ and frequency $f$ is:
$\omega = 2\pi f$
Solving for the frequency $f$:
$f = \frac{\omega}{2\pi}$
Substituting $\omega = 176 \text{ rad/s}$ and $\pi = \frac{22}{7}$:
$f = \frac{176}{2 \times \frac{22}{7}} = \frac{176 \times 7}{44}$
$f = 4 \times 7 = 28 \text{ Hz}$

The potential energy of a simple harmonic oscillator is given by $U(t) = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t)$.
The potential energy reaches its maximum value when $|\sin(\omega t)| = 1$, which first occurs at $\omega t = \frac{\pi}{2}$.
Substituting the angular frequency $\omega = \frac{2\pi}{T}$ into the equation gives $\left(\frac{2\pi}{T}\right) t = \frac{\pi}{2}$, which simplifies to $t = \frac{T}{4}$.
Given that the time is expressed as $t = \frac{T}{2\beta}$, we equate $\frac{T}{4} = \frac{T}{2\beta}$.
Solving for $\beta$ yields $2\beta = 4$, so $\beta = 2$.

[CORRECT_OPTION: 2]

If we express position $x(t)=A sin \;(\omega t+\varphi )$ then $x_0=A sin \;\varphi$ $\;v_0=A \omega \cos \varphi$ $\;\Rightarrow \tan \varphi =\;\frac{\omega x_0}{v_0}$ $\;A=\sqrt{x_0^{2+}\;\frac{v_0^2}{\omega ^2}}$Hence both position and linear momentum of a particle can be expressed as a function of time if we know initial momentum and position

Here $\omega =\sqrt{\;\frac{k}{m}}$and maximum velocity $V=A \omega =A \sqrt{\;\frac{k}{m}}$So, $\;\frac{V_A}{V_B}=\sqrt{\;\frac{k_1}{k_2}}$

$T=2 \pi \sqrt{\;\frac{1}{g}}$At top of mountain $g ↓, \therefore T ↑$

$\;T=2 \pi \sqrt{\;\frac{ℓ}{g}}$ $\;g=\;\frac{G M}{R^2}$ $\;g^{'}=\;\frac{G 4 M}{(2 R)^2}=g$Reason is true but not correct explanation.

Additional buoyant force $g \rho a^2 x=\sigma a^3 A$ $\;A=\;\frac{\rho }{\sigma } \;\frac{g}{a} x$ $\;T=2 \pi \sqrt{\;\frac{\sigma a}{\rho g}}$Now, $\sigma =\;\frac{10 \times 10^{-3}}{10^{-3}}=10$ $\Rightarrow T\;=2 \pi \sqrt{\;\frac{10 \times 0.1}{10^3 \times 10}}$ $\;=2 \pi \times 10^{-2}$

For a simple pendulum executing small-angle oscillations, the restoring torque gives the angular equation of motion $\alpha = -\frac{g}{l}\,\theta$ where $\alpha$ is the angular acceleration, $g$ is the acceleration due to gravity and $l$ is the length of the pendulum. Case 1: First pendulum Length $= l_1$, angular displacement $= \theta_1$ Therefore $\alpha_1 = -\frac{g}{l_1}\,\theta_1$ $-(1)$ Case 2: Second pendulum Length $= l_2$, angular displacement $= \theta_2$ Therefore $\alpha_2 = -\frac{g}{l_2}\,\theta_2$ $-(2)$ The question states that both pendulums have the same angular acceleration, so $\alpha_1 = \alpha_2$ $-(3)$ Substituting $(1)$ and $(2)$ into $(3)$: $-\frac{g}{l_1}\,\theta_1 = -\frac{g}{l_2}\,\theta_2$ The factor $-g$ is common on both sides and cancels out, giving $\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2}$ Cross-multiplying, we obtain the required relation $\theta_1\,l_2 = \theta_2\,l_1$ This matches Option D. Answer : Option D

Energy conservation$\;\frac{1}{2} k(1)^2=\;\frac{1}{2} m v^{2+}\;\frac{1}{2} k(2-x)^2$Compression in the spring $=(2-x)$ $\Rightarrow \;\frac{1}{2} k (1-(2-x)^2) =\;\frac{1}{2} m v^2$ \Rightarrow v= \[100 (1-(2-x)^2) ]$ ^{1 / 2}$$;v=10[1-(2-x)]^{1 / 2}$

$\;\omega =5$ $\; A\;=\sqrt{({\sqrt{7}})^{2+}(2 {\sqrt{7}})^{2+}2 \times 2 \times 7 \cos (\;\frac{\pi }{3}) }$ $\;=7$ $\;a_{\max }$

$x(t)=x_0 sin \;^2 (\;\frac{t}{2}) =\;\frac{x_0}{2}(1-\cos t)$ $\;\text{ Clearly }\; \;\frac{x_0}{2} \;\text{ is mean position. }\;$

A. Assuming no slipping, $T=2 \pi \sqrt{\;{m_{\;\text{total }\;}}/{k}}$ A is correct.B. Assuming no slipping, $a=\;\frac{|F|}{m}$ B is correct.C. $f=(m)(a)=\;\frac{m \times k x}{m+M}$C is correct.D. For no slipping $\;\frac{k x_0}{m+M} \le \mu g$D is correct.E. $\;\; f_{\max }=\mu m g$ $E$ is incorrect.

$\;T=2 s$ $\;A=1 cm$In 12.5 s 6 total oscillation and 1 quarter oscillation.So,$\;D=(6 \times 4 A)+A=25 A$ $\;d=A$ $\;\frac{D}{d}=25$

From phasor diagram particle has to move from ${\text{P}}$ to ${\text{Q}}$ in a circle of radius equal to amplitude of SHM. $\;\cos \phi =\;{\;{{\sqrt{3}} {\text{A}}}/{2}}/{ {\text{A}}}=\;{{\sqrt{3}}}/{2}$ $\;\phi =\;\frac{\pi }{6}$Now, $\;\frac{\pi }{6}=\omega {\text{t}}$ $\;\;\frac{\pi }{6}=\;\frac{2 \pi }{T} t$ $\;\;\frac{\pi }{6}=\;\frac{2 \pi }{6 \pi } t$ $t=\;\frac{\pi }{2}$So, $x=2$

$\;x=4 {\text{m}}, V=2 {\text{m}} / {\ s}, {\text{a}}=16 {\text{m}} / {\ s}^2$ $\;| {\text{a}}|=\omega ^2 x$ $\;\Rightarrow 16=\omega ^2(4)$ $\;\omega =2 {\text{rad}} / {\ s}$ $\;v=\omega {\sqrt{A^{2-}x^2}}$ $\;A=\sqrt{\;\frac{v^2}{\omega ^2}+x^2} \Rightarrow A=\sqrt{\;\frac{4}{4}+16}$ $\;A={\sqrt{17}} {\text{m}}$