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

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

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

If the displacement x and the velocity v of a particle executing simple harmonic motion are related through the expression $4v^2=25-x^2$,then its time period will be:

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

A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resulting amplitude is equal to the amplitude of individual motions, the phase difference between them will be:

1.  $\frac{\mathrm{\pi }}{3}$

2.  $\frac{2\mathrm{\pi }}{3}$

3.  $\frac{\mathrm{\pi }}{6}$

4.  $\frac{\mathrm{\pi }}{2}$

A body oscillates with SHM according to the equation (in SI units), x = 5 cos [2π t + π/4]. At t = 1.5 s, acceleration of the body will be:

A particle of mass \(m\) and charge \(\text-q\) moves diametrically through a uniformly charged sphere of radius \(R\) with total charge \(Q\). The angular frequency of the particle's simple harmonic motion, if its amplitude \(<R\), is given by:
1. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR} }\)
2. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR^2} }\)
3. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR^3}}\)
4. \( \sqrt{\dfrac{m}{4 \pi \varepsilon_0 ~qQ} }\)