Question

A small body of the mass 0.10kg is executing simple harmonic motion whose amplitude is 1.0m and the period is 0.20s. What is the maximum value of the force acting on it? If the oscillations are performed by the means of a spring then what will be the force constant of the spring?

Answer 1

Hint: In order to find max force we have to find maximum acceleration by using the formula for the maximum acceleration for the simple harmonic motion, by finding max acceleration we can find max force and by using this force we can find force constant.
Formula used:
$\begin{align}
  & F=ma \\
 & {{a}_{\max }}=A{{\omega }^{2}} \\
 & \omega =\dfrac{2\pi }{T} \\
\end{align}$

Complete answer:
Given data:
Mass of the body = m = 0.10kg
Amplitude = A = 1.0m
Time period = T = 0.20s
Now to find max force we can use the formula of the force given as,
${{F}_{\max }}=m{{a}_{\max }}...\left( 1 \right)$
Where, m = mass
${{F}_{\max }}$= max. Force
${{a}_{\max }}$= max acceleration of a body.
Here, we don’t know the value of max acceleration to find it we can use the below equation,
${{a}_{\max }}=A{{\omega }^{2}}....\left( 2 \right)$
Where, A = amplitude
ω = angular velocity
We know that formula for the angular velocity is,
$\omega =\dfrac{2\pi }{T}....\left( 3 \right)$
Where, T = time period
Substitute value of the equation (3) in the equation (2)
${{a}_{\max }}=A\times {{\left( \dfrac{2\pi }{T} \right)}^{2}}....\left( 4 \right)$
Now from the equation (1) and (4) max Force is,
${{F}_{\max }}=m\times A\times {{\left( \dfrac{2\pi }{T} \right)}^{2}}...\left( 5 \right)$
Now substitute all the values in the equation (5)
$\begin{align}
  & {{F}_{\max }}=0.1\times 1\times {{\left( \dfrac{2\pi }{0.2} \right)}^{2}} \\
 & =0.1\times 985.96 \\
 & =98.596 \\
 & {{F}_{\max }}=98.6N \\
\end{align}$
Now if the oscillation is performed by means of a spring then we can take amplitude (A) as displacement of the spring (x) and max Force (${{F}_{\max }}$) as a spring force to find spring constant (k) we know that formula for the spring force is,
$F=kx....\left( 6 \right)$
Where, F = spring force
k = spring constant
x = displacement.
Now substitute all the values in the equation (6)
$\begin{align}
  & \Rightarrow 98.6=k\times 1 \\
 & \therefore k=98.6N/m \\
\end{align}$
Therefore the value of the spring constant is $98.6N/m$.

Note:
As we know the amplitude is the maximum displacement of the object from the equilibrium position therefore if we think about spring then we can take the value of the amplitude (A) as displacement spring (x) to find spring constant. (k)