A particle in SHM is described by the displacement equation $\mathrm{x}\left(\mathrm{t}\right)=\mathrm{Acos} \left(\mathrm{\omega t}+\mathrm{\theta }\right).$ If the initial position of the particle is 1 cm and its initial velocity is $\mathrm{\pi }$cm/s, what is its amplitude? (The angular frequency of the particle is $\pi s^{-1}$)

1. 1 cm       
2. $\sqrt{2}$ cm
3.  2 cm     
4. 2.5 cm

If the length of a pendulum is made \(9\) times and mass of the bob is made \(4\) times, then the value of time period will become:
1. \(3T\)
2. \(\dfrac{3}{2}T\)
3. \(4T\)
4. \(2T\)

A simple harmonic wave having an amplitude a and time period T is represented by the equation $y=5 \mathrm{sin\pi }\left(t+4\right)$m Then the value of amplitude (a) in (m) and time period  (T) in second are       

1.   $a=10, T=2$   

2. $a=5, T=1$

3.    $a=10, T=1$    

4. $a=5, T=2$

The period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with acceleration of g/3 then the time period of the pendulum is

1. $\frac{\mathrm{T}}{\sqrt{3}}$

2. $\frac{\mathrm{T}}{3}$

3. $\frac{\sqrt{3}}{2}\mathrm{T}$

4. $\sqrt{3}\mathrm{T}$

The time period of a simple pendulum of length L as measured in an elevator descending with acceleration $\frac{\mathrm{g}}{3}$ is

1. $2\mathrm{\pi }\sqrt{\frac{3\mathrm{L}}{\mathrm{g}}}$

2. $\mathrm{\pi }\sqrt{\left(\frac{3\mathrm{L}}{\mathrm{g}}\right)}$

3. $2\mathrm{\pi }\sqrt{\left(\frac{3\mathrm{L}}{2\mathrm{g}}\right)}$

4. $2\mathrm{\pi }\sqrt{\frac{2\mathrm{L}}{3\mathrm{g}}}$

If the displacement equation of a particle be represented by $y=A\sin Pt+ B\cos Pt$ , the particle executes

1.         A uniform circular motion

2.         A uniform elliptical motion

3.         A S.H.M.

4         A rectilinear motion

A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $\mathrm{Fsin\omega t}$ . If the amplitude of the particle is maximum for $\mathrm{\omega }={\mathrm{\omega }}_1$  and the energy of the particle is maximum for $\mathrm{\omega }={\mathrm{\omega }}_2$, then (where ${\omega }_0$ is natural frequency of oscillation of particle)

1. ${\mathrm{\omega }}_1={\mathrm{\omega }}_0$ and ${\mathrm{\omega }}_2\ne {\mathrm{\omega }}_0$

2. ${\mathrm{\omega }}_1={\mathrm{\omega }}_0$ and ${\mathrm{\omega }}_2={\mathrm{\omega }}_0$

3. ${\mathrm{\omega }}_1\ne {\mathrm{\omega }}_0$ and ${\mathrm{\omega }}_2={\mathrm{\omega }}_0$

4. ${\mathrm{\omega }}_1\ne {\mathrm{\omega }}_0$ and ${\mathrm{\omega }}_2\ne {\mathrm{\omega }}_0$

The displacement of a particle varies according to the relation $x = 4(\mathrm{cos\pi }t + \mathrm{sin\pi }t).$The amplitude of the particle is

1.   8           

2.  – 4

3.   4          

4.   $4\sqrt{2}$