Question

A particle executing SHM maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s². The period of oscillation is(a) π sec (b) π/2 sec(c) 2π sec(d) π/t sec

Answer 1

v = Aw

a = w²A

So,

v/a = 1/w

30/60 = T/2pi

1/2 = T/2pi

T = pi.

Hence, option (a).

Answer 2

The period of oscillation is $\pi\ sec$.

Explanation:

The maximum speed of the particle, v = 30 cm/s

The maximum acceleration of the particle, $a=60\ cm/s^2$

The maximum speed and the acceleration of the particle is given by :

$v=A\omega$........(1)

$a=A\omega^2$.............(2)

From equation (1) and (2) we get :

$\dfrac{v}{a}=\dfrac{1}{\omega}$

Since, $\omega=\dfrac{2\pi}{T}$

$\dfrac{v}{a}=\dfrac{T}{2\pi}$

$T=2\pi (\dfrac{v}{a})$

$T=2\pi (\dfrac{30}{60})$

$T=\pi\ sec$

So, the period of oscillation is $\pi\ sec$. Hence, this is the required solution.

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