Question 21:
moderateA simple harmonic oscillator has an amplitude $A$ and time period $T$. The time required by it to travel from $X = A$ to $X = A/2$ is:
(1992)
Using equation for SHM starting from extreme position: $x = A\cos(\omega t)$.\nSubstitute $x = A/2$: $A/2 = A\cos(2\pi t/T) \implies \cos(2\pi t/T) = 1/2$.\n$2\pi t/T = \pi/3 \implies t = T/6$.
Question 22:
moderateA spring is stretched by $5 \text{ cm}$ by a force $10 \text{ N}$. The time period of the oscillations when a mass of $2 \text{ kg}$ is suspended by it is:
(2021)
$k = F/x = 10 / 0.05 = 200 \text{ N/m}$. Time period $T = 2\pi \sqrt{m/k} = 2\pi \sqrt{2/200} = \frac{2\pi}{10} = 0.628 \text{ s}$.
Question 23:
moderateA pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is $20 \text{ m/s}^2$ at a distance of $5 \text{ m}$ from the mean position. The time period of oscillation is:
(2018)
Acceleration $a = \omega^2 x \Rightarrow 20 = \omega^2 (5) \Rightarrow \omega^2 = 4 \Rightarrow \omega = 2 \text{ rad/s}$. Time period $T = \frac{2\pi}{\omega} = \pi \text{ s}$.
Question 24:
moderateThe angular velocity and the amplitude of a simple pendulum is $ \omega $ and a respectively. At a displacement $ x $ from the mean position if its kinetic energy is $ T $ and potential energy is $ V $, then the ratio of $ T $ to $ V $ is:
(1991)
Kinetic energy is $ T = \frac{1}{2}m\omega^2(a^2 - x^2) $ and potential energy is $ V = \frac{1}{2}m\omega^2x^2 $. Taking the ratio gives $ \frac{T}{V} = \frac{a^2 - x^2}{x^2} $.