Question

A block of mass $2kg$ executes simple harmonic motion under the restoring force of a spring. The amplitude and the time period of motions are $0.2cm\text{and}2\pi sec$ respectively. Find the maximum force exerted by the spring on the block.

• A. $0.002N$
• B. $0.05N$
• C. $0.003N$
• D. $0.004N$

Answer 1

The correct option is D $0.004N$

We know that: $F_{max}=m{w}^{2}A$
Given mass of the block $m=2kg,$
Amplitude $A=0.2cm$,
Time period $T=2\pi sec=6.28sec$
The maximum force exerted will be at extreme positions,
$F_{max}=m{w}^{2}A\dots \left(i\right)$
${}^{\prime }{w}^{\prime }$ can be expressed as below formula
$w=\sqrt{\frac{K}{m}}$
Here, $K$ be the spring constant and $m$ be the mass.
Substitute the value of $w$ in equation $\left(i\right)$,
$F_{max}=m{\left(\sqrt{\frac{K}{m}}\right)}^{2}A$
Substituting the values, we get
$F_{max}=2\times {\left(\frac{2\pi }{6.28}\right)}^2\times (0.2\times {10}^{-2})$
$F_{max}=2\times (1{)}^2\times (0.2\times {10}^{-2})$
$F_{max}=0.004N$
Final answer: (d)