A spring is having a spring constant k. It is cut into two parts A and B whose lengths are in the ratio of m:1. The spring constant of part A will be

1. $\frac{k}{m}$

2. $\frac{k}{m+1}$

3. k

4. $\frac{k(m+1)}{m}$

The displacement-time graph of a particle executing SHM is shown in the figure. Its displacement equation will be: (Time period = 2 second)

1. $\mathrm{x}=10$ $\sin \left(\mathrm{\pi t}+\frac{\mathrm{\pi }}{6}\right)$

2. $\mathrm{x}=10$ $\sin \left(\mathrm{\pi t}\right)$

3. $\mathrm{x}=10$ $\cos \left(\mathrm{\pi t}\right)$

4. $x=5$ $\sin \left(\mathrm{\pi t}+\frac{\mathrm{\pi }}{6}\right)$

All the surfaces are smooth and the system, given below, is oscillating with an amplitude \(\mathrm{A}.\) What is the extension of spring having spring constant \(\mathrm{k_1},\) when the block is at the extreme position?
             

The amplitude of a simple harmonic oscillator is \(A\) and speed at the mean position is \(v_0\) .The speed of the oscillator at the position \(x={A \over \sqrt{3}}\) will be: 

In a simple harmonic oscillation, the graph of acceleration against displacement for one complete oscillation will be:
1. an ellipse
2. a circle
3. a parabola
4. a straight line

A particle executing SHM crosses points A and B with the same velocity. Having taken 3 s in passing from A to
B, it returns to B after another 3 s. The time period of the SHM will be:

1. 15 s

2. 6 s

3. 12 s

4. 9 s

Acceleration of the particle at $t=\frac{8}{\mathrm{3 }}$s from the given displacement (y) versus time (t) graph will be?
                 

1. $\frac{\sqrt{3}{\mathrm{\pi }}^2}{4}$ $cm/s^2$

2. $-\frac{\sqrt{3}{\mathrm{\pi }}^2}{4}$ $cm/s^2$

3. $-{\mathrm{\pi }}^2$ $\mathrm{cm}/{\mathrm{s}}^2$

4. Zero

The time period of the spring-mass system depends upon:

A simple pendulum attached to the ceiling of a stationary lift has a time period of 1 s. The distance y covered by the lift moving downward varies with time as y = 3.75 ${\mathrm{t}}^2$, where y is in meters and t is in seconds. If g = 10 $\mathrm{m}/{\mathrm{s}}^2$, then the time period of the pendulum will be:

The graph between the velocity (v) of a particle executing S.H.M. and its displacement (x) is shown in the figure. The time period of oscillation for this SHM will be

1.  $\sqrt{\frac{\mathrm{\alpha }}{\mathrm{\beta }}}$

2.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{\alpha }}{\mathrm{\beta }}}$

3.  $2\mathrm{\pi }\left(\frac{\mathrm{\beta }}{\mathrm{\alpha }}\right)$

4.  $2\mathrm{\pi }\left(\frac{\mathrm{\alpha }}{\mathrm{\beta }}\right)$