Question

A particle executing simple harmonic motion of amplitude 5cm has maximum speed of 31. 4 cm / s. The frequency of its ossillation is​

Answer 1

Explanation:

Maximum speed of a particle executing SHM is given by,

vmax = a ω = a ( 2 πn ) => n = vmax / 2πa

Where,

a = amplitude of oscillation

n = frequency of oscillation

Here, vmax = 31.4 cm/s, a = 5cm

Substituting, the given values we have

n = 31.4 / 2 × 3.14 × 5 = 1 Hz!

∴The frequency of its ossillation is 1 Hz!

Answer 2

Answer:

1Hz

V

max

=aω⇒f=

2π

ω

=

2πa

V

max

=

2π×5

31.4

=

10π

31.4

=1Hz

Explanation:

hope it's helpful