Question

a particle vibrates in shm along a straight line. its greatest acceleration is 5π2 cm/s2 and when its distance from the equilibrium position is 4cm the velocity of the particle is 3πcm/s then the period of oscillation​

Answer 1

Period of oscillation will be 0.636 sec

Explanation:

We have given that maximum acceleration in SHM is $\alpha =4\pi ^2=4\times 3.14^2=39.44cm/sec^2$

Displacement from the equilibrium position x=4 cm

We know that in SHM maximum acceleration is given by

$\alpha =\omega ^2x$

$39.44=\omega ^2\times 4$

$\omega ^2=9.86rad/sec$

We know that time period is given by

$T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{9.86}=0.636sec$

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