Question

a particle executing SHM completes 1200 oscillation per minute and passes through the mean position with a velocity of 31.4m/s determine the maximum displacement of the particle from the mean position. Also obtain the displacement equation of the particle if its displacement be zero at the instant t=0

Answer 1

number of oscillation per minute = 1200
so, number of oscillation per second = 1200/60 = 20
hence, frequency , f= number of oscillation per second = 20 Hz.

angular frequency, $\omega$ = 2πf
= 2π × 20 = 40π rad/s

we know, $\bf{v=\omega r}$

31.4 = 40π × r
31.4 = 40 × 3.14 × r
r = 0.25m or 25cm

hence, displacement of the particle from mean position is 25cm.

now, equation of SHM can be written as
$x=Asin\omega t$
x { in cm} = 25sin(40π)t

Answer 2

Given:

Oscillations = 1200 per minute

Velocity = 31.4 m / s

To find:

• The maximum displacement
• Displacement equation if displacement and time are 0

Solution:

The number of oscillations per second = 1200 / 60

Hence, 20 oscillations per second

By formula,

Angular frequency = 2 * π * f

Substituting,

We get,

2 * π * 20

Angular frequency = 40 rad / s

We also know that,

Velocity = Angular frequency * Distance

Distance = Velocity / Angular frequency

Substituting,

31.4 / 40

We get,

Distance = 0.25 m

The general equation of SHM,

x = A sinω t

Substituting,

x = 25 sin ( 45 π ) t