Question

Question: A0.12 kgbody undergoes simple harmonic motion of amplitude 8.5 cmand period20 s.

1. What is the magnitude of the maximum force acting on it?
2. If the oscillations are produced by a spring, what is the spring constant?

Short answer

Answer:

1. Magnitude of maximum force is 10.1N
2. Spring constant if force is produced by the spring is 120 N/m

Step-by-step solution

The given data

1. Mass of the object, m= 12.0 Kg
2. Amplitude of motion, A = 8.5 cm
3. Period of motion, T = 0.2 s

Understanding the concept of Newton’s laws of motion

According to Newton’s second law, force is a product of mass and acceleration. Hence, we can say force is maximum when acceleration is maximum and the value of acceleration is maximum when the object is at the end position.

Formula:

The restoring force is proportional to displacement,

$F=m{\omega }^2{X}_m$ (i)

The angular frequency of a body in simple harmonic motion,

$\omega =\sqrt{\frac{\text{k}}{\text{m}}}$(ii)

Calculation of magnitude of maximum force

The angular frequency of a body in simple harmonic motion is given as:

$$\begin{gathered} \omega =\frac{2\pi }{T} \\ =\frac{2\pi }{0.2} \\ =31.41\text{rad}/\text{sec} \end{gathered}$$

Now, using equation (i) and the given values, we can get the force as:

$$\begin{gathered} F=0.12\times 31.{41}^2\times 0.085 \\ =10.1\text{N} \end{gathered}$$

Hence, the magnitude of force is 10.1 N

b) Calculation of the value of spring constant  

Using equation (ii) and the given values, the spring constant can be given as:

$$\begin{gathered} 31.41=\sqrt{\frac{\text{k}}{0.12}} \\ \text{k}=118.4\text{N}/\text{m} \\ \simeq 120\text{N}/\text{m} \end{gathered}$$

Hence, the value of the spring constant is120 N/m.