If the length of a simple pendulum is increased by 2%, then the time period: (1997)
Time period $ T \propto l^{1/2} $. For small percentage changes, $ \frac{\Delta T}{T} \times 100 = \frac{1}{2} \left( \frac{\Delta l}{l} \times 100 \right) $. Thus, the time period increases by $ \frac{1}{2} \times 2\% = 1\% $.
A body of mass 5 kg hangs from a spring and oscillates with a time period of $ 2\pi $ seconds. If the ball is removed, the length of the spring will decrease by: (1994)
Given $ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi $ sec, we get $ m/k = 1 $. The original extension was $ x = \frac{mg}{k} $. Substituting $ m/k = 1 $ gives $ x = g $ metres. The spring decreases by this amount.
A seconds pendulum is mounted in a rocket. Its period of oscillation will decrease when rocket is: (1994)
Time period $ T = 2\pi\sqrt{\frac{l}{g_{eff}}} $. To decrease the period, $ g_{eff} $ must increase. Moving upward with uniform acceleration $ a $ gives $ g_{eff} = g + a $, which increases effective gravity and decreases the period.
In case of a forced vibration, the resonance wave becomes very sharp when the: (2003)
The sharpness of resonance in a forced oscillator is inversely proportional to the damping present in the system. Therefore, the resonance wave becomes very sharp when the damping force is small.
When an oscillator completes 100 oscillation its amplitude reduced to $\frac{1}{3}$ of initial value. What will be its amplitude, oscillation when it completes 200 oscillation? (2002)
Amplitude after $n$ oscillations is given by $A = A_0 e^{-k n}$. For $n = 100$, $A_{100} = A_0 e^{-100k} = A_0 \frac{1}{3}$. For $n = 200$, $A_{200} = A_0 e^{-200k} = A_0 (e^{-100k})^2 = A_0 (\frac{1}{3})^2 = A_0 \frac{1}{9}$.
The amplitude of a S.H.M. reduces to $1/3$ in first $20 \text{ secs}$, then in first $40 \text{ sec.}$ its amplitude becomes: (1999)
In damped S.H.M., amplitude at time $t$ is $A(t) = A_0 e^{-bt}$. At $t = 20 \text{ s}$, $A(20) = A_0 e^{-20b} = \frac{A_0}{3}$. At $t = 40 \text{ s}$, $A(40) = A_0 e^{-40b} = A_0 (e^{-20b})^2 = A_0 (\frac{1}{3})^2 = \frac{A_0}{9}$.