The equation of an SHM is given as y=3sinωt + 4cosωt where y is in centimeters. The amplitude of the SHM will be?

A particle executing simple harmonic motion of amplitude \(5~\text{cm}\) has a maximum speed of \(31.4~\text{cm/s}.\) The frequency of its oscillation will be:
1. \(1~\text{Hz}\)
2. \(3~\text{Hz}\)
3. \(2~\text{Hz}\)
4. \(4~\text{Hz}\)

Two simple harmonic motions of angular frequency 100 rad s ^-1 and 1000 rad $s^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration will be:
1. 1:10
2. 1:10²
3. 1:10³                               
4. 1:10⁴

If a particle is executing SHM, with an amplitude A, the distance moved and the displacement of the body in a time equal to its time period are, respectively:

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: 

A body performs simple harmonic motion about x=0 with an amplitude a and a time period T. The speed of the body at $\mathrm{x}=\frac{\mathrm{a}}{2}$ will be:
1. $\frac{\mathrm{\pi a}\sqrt{3}}{2\mathrm{T}}$                                         
2. $\frac{\mathrm{\pi a}}{\mathrm{T}}$
3. $\frac{3{\mathrm{\pi }}^2\mathrm{a}}{\mathrm{T}}$                                           
4. $\frac{\mathrm{\pi a}\sqrt{3}}{\mathrm{T}}$

A particle executes SHM with a time period of 4 s. The time taken by the particle to go directly from its mean position to half of its amplitude will be:

1.  $\frac{1}{3}$ s

2.  1 s

3.  $\frac{1}{2}$ s

4.   2 s

A point performs simple harmonic oscillation of period \(\mathrm{T}\) and the equation of motion is given by; \(x=a \sin (\omega t+\pi / 6)\). After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?
1. \( \frac{T}{8} \)

2. \( \frac{T}{6} \)

3. \(\frac{T}{3} \)

4. \( \frac{T}{12}\)

The amplitude and the time period in an S.H.M. are 0.5 cm and 0.4 sec respectively. If the initial phase is $\mathrm{\pi }/2$ radian, then the equation of S.H.M. will be:

1. $\mathrm{y}=0.5\sin 5\mathrm{\pi t}$

2. $\mathrm{y}=0.5\sin 4\mathrm{\pi t}$

3. $\mathrm{y}=0.5\sin 2.5\mathrm{\pi t}$

4. $\mathrm{y}=0.5\cos 5\mathrm{\pi t}$

A particle executes linear SHM between \(x=A.\) The time taken to go from \(0\) to \(A/2\) is ${\mathrm{T}}_1$ and to go from \(A/2\) to \(A\) is ${\mathrm{T}}_2$, then: