Question

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s.
(a) What is the magnitude of the maximum force acting on it?

Answer 1

Explanation:

The maximum force acting on body executing SHM can be calculated by,

• $\sf F = mA\omega^2$

where,

• m is mass of the body
• A is amplitude
• $\sf \omega$ is angular frequency

We are given : A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s.

• Here Mass of the body, m = 0.12 kg
• amplitude,A = 8.5 cm = 0.085 m
• Time period, T= 0.20 s

First we need to calculate the angular frequency. Angular frequency (ω) is calculated by,

$\sf \omega = \dfrac {2π}{T}$

$\sf \omega = \dfrac {2 \times 3.14}{0.20}$

$\sf \omega = \dfrac{6.28}{0.20}$

$\sf \omega = 31.4 \ s^{-1}$

Now,

$\sf F = mA\omega^2$

$\sf F = 0.12 \times 0.085 \times (31.4)^2$

$\sf F = 0.12 \times 0.085 \times 985.96$

$\sf F = 10.05 N$

Hence, the magnitude of the maximum force acting on it is 10.05 N

Answer 2

Final answer:

The magnitude of the maximum force acting on the body during simple harmonic motion is 0.85 N.

Explanation:

To find the magnitude of the maximum force acting on the body, we can use the equation:

F = kA

where F is the force, k is the force constant, and A is the amplitude of the oscillations.

Using the given values, the force constant is 10.0 N/m and the amplitude is 8.5 cm (which is equal to 0.085 m).

F = (10.0 N/m) * (0.085 m) = 0.85 N

Therefore, the magnitude of the maximum force acting on the body is 0.85 N.