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The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10\, cm$ to $8\, cm$ in $40\, seconds$ . Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$ . The time in which amplitude of this pendulum will reduce from $10\, cm$ to $5\, cm$ in carbon dioxide will be close to ..... $s$ $(ln\, 5 = 1.601,ln\, 2 = 0 .693)$
The instantaneous displacement of a simple pendulum oscillator is given by $x = A\cos \left( {\omega t + \frac{\pi }{4}} \right)$. Its speed will be maximum at time
The amplitude of vibration of a particle is given by ${a_m} = ({a_0})/(a{\omega ^2} - b\omega + c);$ where ${a_0},a,b$ and $c$ are positive. The condition for a single resonant frequency is
A simple pendulum of length $ l$ has a brass bob attached at its lower end. Its period is $T$. If a steel bob of same size, having density $ x$ times that of brass, replaces the brass bob and its length is changed so that period becomes $2T$, then new length is
Two bodies performing $SHM$ have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is