Question

The kinetic energy of a particle executing simple harmonic motion at a displacement of $3~\text{cm$ from the mean position is $4~\text{mJ}$. If the amplitude of the particle is $5~\text{cm}$, then the maximum force acting on the particle is}

• A. $0.25~\text{N}$
• B. $0.50~\text{N}$
• C. $0.75~\text{N}$
• D. $1.25~\text{N}$

Hint:

In SHM, use $K = \frac{1}{2}k(A^2 - x^2)$ to find spring constant, and then use $F = kA$ for max force.

Answer 1

The Correct Option is A

Kinetic energy in SHM is given by: $K = \dfrac{1}{2}k(A^2 - x^2)$
Given $K = 4 \times 10^{-3}~\text{J}$, $x = 0.03~\text{m}$, $A = 0.05~\text{m}$
Substitute: $4 \times 10^{-3} = \dfrac{1}{2}k(0.0025 - 0.0009) = \dfrac{1}{2}k(0.0016)$
$\Rightarrow k = \dfrac{4 \times 10^{-3} \times 2}{0.0016} = 5~\text{N/m}$
Maximum force $F = kA = 5 \cdot 0.05 = 0.25~\text{N}$