The graph between the velocity \((v)\) of a particle executing SHM and its displacement \((x)\) is shown in the figure. The time period of oscillation for this SHM will be:
1. \(\sqrt{\frac{\alpha}{\beta}}\)
2. \(2\pi\sqrt{\frac{\alpha}{\beta}}\)
3. \(2\pi\left(\frac{\beta}{\alpha}\right)\)
4. \(2\pi\left(\frac{\alpha}{\beta}\right)\)
Acceleration-time (\(a\text-t\)) graph for a particle performing SHM is shown in the figure. Select the incorrect statement.
1. | Displacement of a particle at \(A\) is negative. |
2. | The potential energy of the particle at \(C\) is minimum. |
3. | The velocity of the particle at \(B\) is positive. |
4. | Speed of particle at \(D\) is decreasing. |
A simple pendulum is pushed slightly from its equilibrium towards the left and then set free to execute the simple harmonic motion. Select the correct graph between its velocity (\(v\)) and displacement (\(x \)).
1. | 2. | ||
3. | 4. |
1. | maybe \(K_0\) |
2. | must be \(K_0\) |
3. | may be more than \(K_{0}\) |
4. | both (1) and (3) |
1. | the gravity of the earth | 2. | the mass of the block |
3. | spring constant | 4. | both (2) & (3) |
Acceleration of the particle at \(t = \frac{8}{3}~\text{s}\) from the given displacement \((y)\) versus time \((t)\) graph will be?
1. \(\frac{\sqrt{3}\pi^2}{4}~\text{cm/s}^2\)
2. \(-\frac{\sqrt{3}\pi^2}{4}~\text{cm/s}^2\)
3. \(-\pi^2~\text{cm/s}^2\)
4. zero
1. | \(T_1<T_2\) | 2. | \(T_1>T_2\) |
3. | \(T_1=T_2\) |
4. | \(T_1= 2T_2\) |
A spring-block system oscillates with a time period \(T\) on the earth's surface. When the system is brought into a deep mine, the time period of oscillation becomes \(T'.\) Then one can conclude that:
1. \(T'>T\)
2. \(T'<T\)
3. \(T'=T\)
4. \(T'=2T\)