A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector \(\vec a\) is correctly shown in:
1. | 2. | ||
3. | 4. |
A block \(P\) of mass \(m\) is placed on a frictionless horizontal surface. Another block \(Q\) of same mass is kept on \(P\) and connected to the wall with the help of a spring of spring constant \(k\) as shown in the figure. \(\mu_s\) is the coefficient of friction between \(P\) and \(Q\). The blocks move together performing SHM of amplitude \(A\). The maximum value of the friction force between \(P\) and \(Q\) will be:
1. \(kA\)
2. \(\frac{kA}{2}\)
3. zero
4. \(\mu_s mg\)
1. | simple harmonic motion of frequency \(\frac{\omega}{\pi}\). |
2. | simple harmonic motion of frequency \(\frac{3\omega}{2\pi}\). |
3. | non-simple harmonic motion. |
4. | simple harmonic motion of frequency \(\frac{\omega}{2\pi}\). |
Two 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 acceleration will be:
1. \(1:10\)
2. \(1:10^{2}\)
3. \(1:10^{3}\)
4. \(1:10^{4}\)
1. | \(\pi \) | 2. | \(2 \pi \) |
3. | \(4 \pi \) | 4. | \(6 \pi\) |
The time period of a spring mass system at the surface of earth is 2 second. What will be the time period of this system on the moon where acceleration due to gravity is of the value of g on earth's surface?
1. | \(\frac{1}{\sqrt{6}} ~\mathrm{seconds} \) | 2. | \(2 \sqrt{6}~ \mathrm{seconds} \) |
3. | \(2~ \mathrm{seconds} \) | 4. | \( 12~\mathrm{ seconds}\) |
1. | \( \dfrac{T}{12} \) | 2. | \(\dfrac{5 T}{12} \) |
3. | \( \dfrac{7 T}{12} \) | 4. | \(\dfrac{2 T}{3}\) |