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

(1)$\frac{T}{8}$

(2) $\frac{T}{6}$

(3) $\frac{T}{3}$

(4) $\frac{T}{12}$

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. \({2a \pi \over T}\)  

Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is -

(1)   1 : 2                 

(2)  2 : 1

(3)   2 : 3                 

(4)  3 : 2

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 same mass and maximum velocity, then

(a) $A_1{\omega }_1=A_2{\omega }_2=A_3{\omega }_3$        (b)  $A_1{{\omega }_1}^2=A_2{{\omega }_2}^2=A_3{A_3}^2$

(b) ${A_1}^2{\omega }_1={A_2}^2{\omega }_2={A_3}^2{\omega }_3$   (d) ${A_1}^2{{\omega }_1}^2={A_2}^2{{\omega }_2}^2=A^2$  

The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3}\mathrm{cm}/\sec$ will be

(1)         $2\sqrt{3}cm$             

(2)  $\sqrt{3}cm$

(3)         1 cm                   

(4)  2 cm

The maximum velocity of a simple harmonic motion represented by $y=3 \sin \left(100t+\frac{\mathrm{\pi }}{6}\right)$ is given by

(1)       300       

(2)   $\frac{3\mathrm{\pi }}{6}$

(3)      100         

(4)   $\frac{\mathrm{\pi }}{6}$

The displacement equation of a particle is $x=3\sin 2t+4\cos 2t$ The amplitude and maximum velocity will be respectively

(a) 5, 10      

(b) 3, 2

(c) 4, 2       

(d) 3, 4

The instantaneous displacement of a simple pendulum oscillator is given by $x=A \cos \left(\omega t+\frac{\mathrm{\pi }}{4}\right)$ . Its speed will be maximum at time

(1) $\frac{\mathrm{\pi }}{4\omega }$      

(2) $\frac{\mathrm{\pi }}{2\omega }$

(3) $\frac{\mathrm{\pi }}{\omega }$                

(4) $\frac{2\mathrm{\pi }}{\omega }$

The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

(a)      $144{\mathrm{\pi }}^2\mathrm{m}/{\sec }^2$        (b)     $144m/se{c}^2$

c)      $\frac{144}{{\mathrm{\pi }}^2}m/se{c}^2$           (d)     $288{\mathrm{\pi }}^2\mathrm{m}/{\sec }^2$   

A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by

(1)   – A Kx           

(2)   A cos (Kx)

(3)   A exp (– Kx) 

(4)   A Kx