Question 31:
moderateA block is resting on a piston which is moving vertically executing SHM of period 1 s. At what minimum amplitude of motion, will the block and piston separate? (take \(\pi^2 = 10\))
Separation occurs when the maximum downward acceleration of the piston equals \(g\). Thus, \(\omega^2 A = g \implies \left(\frac{2\pi}{T}\right)^2 A = g \implies 4\pi^2 A = 10 \implies 40 A = 10 \implies A = 0.25\text{ m}\).
Question 32:
moderateIf \( x = 2\sin\left(\frac{\pi}{2}t\right) \) represents the motion of a particle executing SHM, the maximum speed of the particle in \( \text{m s}^{-1} \) is (All parameters are in SI units)
Comparing the given equation with the standard SHM equation \( x = A\sin(\omega t) \), we get \( A = 2\text{ m} \) and \( \omega = \frac{\pi}{2}\text{ rad/s} \). The maximum speed is \( v_{\max} = A\omega = 2 \times \frac{\pi}{2} = \pi\text{ m/s} \).
Question 33:
moderateFor the damped oscillator, if time taken for its amplitude of vibrations to drop to half of its initial value is \(T\) then time taken for amplitude to drop to one eighth amplitude is
Amplitude decays exponentially as \(A = A_0 e^{-\gamma t}\). Since it halves in time \(T\), to drop to \(1/8 = (1/2)^3\) of its initial value, it takes exactly \(3T\).
Question 34:
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 35:
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 36:
moderateTwo particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to paths of the two particles. The phase difference is:
(2011 Mains)
Let displacement be $x = A\sin(\omega t + \phi)$. When they cross, $x = A/2$. $A/2 = A\sin(\phi) \Rightarrow \sin(\phi) = 1/2$. The two phases are $\pi/6$ and $5\pi/6$ (since moving in opposite directions). Phase difference $$= 5\pi/6 - \pi/6 = 4\pi/6 = 2\pi/3$$.
Question 37:
moderateThe displacement of a particle along the x-axis is given by $x = a\sin^2\omega t$. The motion of the particle corresponds to:
(2010 Pre)
Equation is $x = a\sin^2\omega t = \frac{a}{2}(1 - \cos 2\omega t)$. This represents SHM about the mean position $x = a/2$. The angular frequency is $2\omega$. The frequency is $f = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}$.
Question 38:
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, $T \propto \frac{1}{\sqrt{A}}$.
Question 39:
moderateThe composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi$ results in the displacement of the particle along:
(1990)
Let $x = A\sin(\omega t)$ and $y = B\sin(\omega t + \pi) = -B\sin(\omega t)$.\nThen $y/x = -B/A \implies y = -(B/A)x$.\nThis represents the equation of a straight line.
Question 40:
moderateThe particle executing simple harmonic motion has a kinetic energy $K_0 \cos^2 \omega t$. The maximum values of the potential energy and the total energy are respectively:
(2007)
The maximum kinetic energy is $K_0$.\nIn an ideal SHM without damping, the total energy remains conserved and equals the maximum kinetic energy, which is $K_0$.\nThe maximum potential energy is also equal to the total energy, which is $K_0$.