The angular velocities of three bodies in simple harmonic motion are ${\omega }_1,$ ${\omega }_2,$ ${\omega }_3$ with their respective amplitudes as $A_1,$ $A_2,$ $A_3$. If all the three bodies have the same mass and maximum velocity, then:

An SHM has an amplitude \(a\) and  a time period \(T.\) The maximum velocity will be:
1. \({4a \over T}\)       
2.  \({2a \over T}\)
3. \({2 \pi \over T}\)
4. \({2a \pi \over T}\)  

The equation of motion of a particle is \({d^2y \over dt^2}+Ky=0 \) where \(K\) is a positive constant. The time period of the motion is given by: 

Displacement versus time curve for a particle executing SHM is shown in the figure. Choose the correct statement/s.

The velocity-time diagram of a harmonic oscillator is shown in the figure given below. The frequency of oscillation will be:
                

1. 25 Hz
2. 50 Hz
3. 12.25 Hz
4. 33.3 Hz

If the time of mean position from amplitude (extreme) position is 6 seconds, then the frequency of SHM will be:

The figure shows the circular motion of a particle. The radius of the circle, the period, the sense of revolution, and the initial position are indicated in the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P will be:

1. $x\left(t\right)=B$ $\sin \left(\frac{2\mathrm{\pi t}}{30}\right)$

2. $x\left(t\right)=B$ $\cos \left(\frac{\mathrm{\pi t}}{15}\right)$

3. $x\left(t\right)=B$ $\sin \left(\frac{\mathrm{\pi t}}{15}+\frac{\mathrm{\pi }}{2}\right)$

4. $x\left(t\right)=B$ $\cos \left(\frac{\mathrm{\pi t}}{15}+\frac{\mathrm{\pi }}{2}\right)$