Question

A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?

Answer 1

1.11 m/s²

Step-by-step explanation:

Amplitude, A = 0.43 m

Time period, T = 3.9 second

Let a be the maximum acceleration and ω be the angular frequency.

ω = 2π/T

ω = ( 2 x 31.4) / 3.9

ω = 1.61 rad/s

Maximum acceleration

a = ω²A

a = 1.61 x 1.61 x 0.43

a = 1.11 m/s²

Answer 2

Maximum acceleration in the simple harmonic motion will be
`0.854rad/sec^2`

Step-by-step explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to
`\omega =(2\pi )/(T)=(2* 3.14)/(3.9)=1.61rad/sec`

Maximum acceleration is given by
`a_(max)=\omega ^2A=1.61^2* 0.43=0.854rad/sec^2`

So maximum acceleration in the simple harmonic motion will be
`0.854rad/sec^2`