Question

A 2.0-kg object is attached to a spring (k = 55.6 N/m) that hangs vertically from the ceiling. The object is displaced 0.045 m vertically. When the object is released, the system undergoes simple harmonic motion. What is the magnitude of the maximum acceleration of the object?

Answer 1

34.78 m/s^2

Step-by-step explanation:

mass of object, m = 2 kg

Spring constant, K = 55.6 N/m

Displacement, A = 0.045 m

Angular velocity

`\omega =\sqrt{(k)/(m)}`

`\omega =\sqrt{(55.6)/(2)}`

ω = 27.8 rad/s

The value of maximum acceleration is given by

a = ω² A

a = 27.8 x 27.8 x 0.045 = 34.78 m/s^2

Thus, the maximum acceleration is 34.78 m/s^2.

Answer 2

Maximum acceleration will be
`1.251m/sec^2`

Step-by-step explanation:

We have given mass of the object m = 2 kg

Spring constant k = 55.6 N/m

Amplitude is given as A = 0.045 m

We know that maximum acceleration in SHM is given by

Maximum acceleration
`=A\omega ^2`

We know that
`\omega ^2=(k)/(m)=(55.6)/(2)=27.8`

So maximum acceleration =
`27.8* 0.045=1.251m/sec^2`