Question 1:
easyThe phase difference between displacement and acceleration of a particle in a simple harmonic motion is:
(2020)
Displacement is given by $ y = A \sin(\omega t) $ and acceleration is $ a = -A\omega^2 \sin(\omega t) = A\omega^2 \sin(\omega t + \pi) $. Thus, the phase difference is $ \pi \text{ rad} $.
Question 2:
easyIdentify the function which represents a periodic motion.
(2020-Covid)
The function $ \sin \omega t + \cos \omega t $ represents a superposition of two simple harmonic motions, making it periodic. The other given functions do not repeat their values over equal intervals of time.
Question 3:
easyAverage velocity of a particle executing SHM in one complete vibration is :
(2019)
In one complete vibration, the particle returns to its starting point, so the net displacement is zero. Since average velocity is total displacement divided by total time, it is zero.
Question 4:
moderateA particle executes linear simple harmonic motion with an amplitude of $ 3 \text{ cm} $. When the particle is at $ 2 \text{ cm} $ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is: (2017-Delhi)
Given $ |v| = |a| $, we have $ \omega \sqrt{A^2 - x^2} = \omega^2 x $. Substituting $ A = 3 $ and $ x = 2 $, we get $ \sqrt{3^2 - 2^2} = \omega (2) \implies \omega = \frac{\sqrt{5}}{2} $. The time period is $ T = \frac{2\pi}{\omega} = \frac{4\pi}{\sqrt{5}} $.
Question 5:
easyWhen two displacements represented by $ y_1 = a \sin(\omega t) $ and $ y_2 = b \cos(\omega t) $ are superimposed, the motion is:
(2015)
The resultant displacement is $ y = y_1 + y_2 = a \sin(\omega t) + b \cos(\omega t) $. This equation represents a single simple harmonic motion with a resultant amplitude of $ R = \sqrt{a^2 + b^2} $.
Question 6:
moderateA particle is executing S.H.M. along a straight line. Its velocities at distances $ x_1 $ and $ x_2 $ from the mean position are $ v_1 $ and $ v_2 $, respectively. Its time period is:
(2015)
Velocity in SHM is $ v^2 = \omega^2(A^2 - x^2) $. So, $ v_1^2 = \omega^2(A^2 - x_1^2) $ and $ v_2^2 = \omega^2(A^2 - x_2^2) $. Subtracting these equations gives $ v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2) $, yielding $$ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} $$.
Question 7:
easyA particle is executing a simple harmonic motion. Its maximum acceleration is $ \alpha $ and maximum velocity is $ \beta $. Then, its time period of vibration will be:
(2015 Re)
Maximum acceleration is $ a_{\text{max}} = A\omega^2 = \alpha $ and maximum velocity is $ v_{\text{max}} = A\omega = \beta $. Dividing them, we get $ \omega = \frac{\alpha}{\beta} $. Thus, the time period is $ T = \frac{2\pi}{\omega} = \frac{2\pi \beta}{\alpha} $.
Question 8:
easyA particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:
(2007)
Equation from equilibrium is $x = a\sin(\omega t)$. For $x = a/2$, $a/2 = a\sin(\omega t) \Rightarrow \sin(\omega t) = 1/2$. This gives $\omega t = \pi/6$. Substituting $\omega = 2\pi/T$, we get $$\frac{2\pi}{T}t = \pi/6 \Rightarrow t = T/12$$.
Question 9:
easyThe phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:
(2007)
In SHM, velocity leads displacement by a phase of $\pi/2$, and acceleration leads velocity by a phase of $\pi/2$ (or $0.5\pi$). Thus, the phase difference between velocity and acceleration is $0.5\pi$.
Question 10:
moderateA rectangular block of mass $m$ and area of cross-section $A$ floats in a liquid of density $\rho$. If it is given a small vertical displacement from equilibrium it undergoes with a time period $T$, then
(2006)
Restoring force on the block is $F = -(\rho A g)y$. Acceleration $a = F/m = -(\frac{\rho A g}{m})y$. This is SHM with $\omega^2 = \frac{\rho A g}{m}$. Time period $T = 2\pi\sqrt{\frac{m}{\rho A g}}$. Therefore, $T \propto \frac{1}{\sqrt{A}}$.