Question 131:
easyA simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $a$ and time period $T$. The speed of pendulum at $x = a/2$ will be:
(2009)
Velocity in SHM is given by $v = \omega\sqrt{A^2 - x^2}$. Here $A = a$, $x = a/2$, and $\omega = \frac{2\pi}{T}$. So, $v = \frac{2\pi}{T}\sqrt{a^2 - \frac{a^2}{4}} = \frac{2\pi}{T} \times \frac{a\sqrt{3}}{2} = \frac{\pi a \sqrt{3}}{T}$.
Question 132:
easyA point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = a \sin(\omega t + \pi/6)$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
(2008)
Velocity $v = \frac{dx}{dt} = a\omega\cos(\omega t + \pi/6)$. Maximum velocity is $a\omega$. Given $v = \frac{a\omega}{2}$, so $\cos(\omega t + \pi/6) = 1/2$. This gives $\omega t + \pi/6 = \pi/3 \Rightarrow \omega t = \pi/6$. Since $\omega = 2\pi/T$, we get $\frac{2\pi t}{T} = \frac{\pi}{6} \Rightarrow t = \frac{T}{12}$.
Question 133:
easyTwo Simple Harmonic Motions of angular frequency $100 \text{ rad s}^{-1}$ and $1000 \text{ rad s}^{-1}$ have the same displacement amplitude. The ratio of their maximum accelerations is:
(2008)
Maximum acceleration in SHM is given by $a_{\text{max}} = \omega^2 A$. Since amplitude $A$ is the same, $a_{\text{max}} \propto \omega^2$. The ratio is $a_1 / a_2 = (\omega_1 / \omega_2)^2 = (100 / 1000)^2 = (1/10)^2 = 1 : 100 = 1 : 10^2$.
Question 134:
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 135:
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 136:
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 137:
easyWhich one of the following statements is true for the speed ‘$v$’ and the acceleration ‘$a$’ of a particle executing simple harmonic motion?
(2004)
In simple harmonic motion, speed is maximum at the mean position.\nAt this mean position, the displacement is zero, causing the restoring force and acceleration to be zero.
Question 138:
moderateThe equations of two waves given as $x = a\cos(\omega t + \delta)$ and $y = a\cos(\omega t + \alpha)$, Where $\delta = \alpha + \pi/2$, then resultant wave represent:
(2000)
Substitute $\delta$: $x = a\cos(\omega t + \alpha + \pi/2) = -a\sin(\omega t + \alpha)$.\nCombining with $y = a\cos(\omega t + \alpha)$ yields $x^2 + y^2 = a^2$, a circle.\nThe direction of motion traces anti-clockwise as time progresses.
Question 139:
easyIf time of mean position from amplitude (extreme) position is $6\text{s}$. Then the frequency of S.H.M. will be:
(1998)
The time taken to travel from the extreme position to the mean position is $T/4$.\nThus, $T/4 = 6 \implies T = 24 \text{ s}$.\nFrequency $f = 1/T = 1/24 \approx 0.04 \text{ Hz}$.
Question 140:
easyTwo S.H.M.s with same amplitude and time period, when acting together in perpendicular directions with a phase difference of $\pi/2$, give rise to:
(1997)
Superposition of two mutually perpendicular SHMs of equal amplitude and a phase difference of $\pi/2$ produces a circular trajectory.\n$x = A\sin(\omega t)$ and $y = A\sin(\omega t + \pi/2) = A\cos(\omega t)$ gives $x^2 + y^2 = A^2$.