Question 171:
easyIf 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\% $.
Question 172:
moderateA 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.
Question 173:
moderateA 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.
Question 174:
easyIn 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.
Question 175:
easyWhen 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 $n = 200$, $A_{200} = A_0 e^{-200k} = A_0 (e^{-100k})^2 = A_0 (\frac{1}{3})^2 = A_0 \frac{1}{9}$.
Question 176:
easyThe 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 $t = 40 \text{ s}$, $A(40) = A_0 e^{-40b} = A_0 (e^{-20b})^2 = A_0 (\frac{1}{3})^2 = \frac{A_0}{9}$.