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

Answer:

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

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 $\omega ^2=\frac{k}{m}=\frac{55.6}{2}=27.8$

So maximum acceleration = $27.8\times 0.045=1.251m/sec^2$

Answer 2

Answer:

34.78 m/s^2

Explanation:

mass of object, m = 2 kg

Spring constant, K = 55.6 N/m

Displacement, A = 0.045 m

Angular velocity

$\omega =\sqrt{\frac{k}{m}}$

$\omega =\sqrt{\frac{55.6}{2}}$

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.