Physics - SHM Question with Solution | TestHub

The amplitude of the motion is $A=10\text{ cm}=0.1\text{ m}$.

The kinetic energy of a block in simple harmonic motion at a distance $x$ from the equilibrium position is given by $K=\frac{1}{2}k(A^2-x^2)$.

Given $K=0.25\text{ J}$ at $x=5\text{ cm}=0.05\text{ m}$.

Substituting the values:

$0.25=\frac{1}{2}k((0.1{)}^2-(0.05{)}^2)$

$0.5=k(0.01-0.0025)$

$0.5=k(0.0075)$

$k=\frac{0.5}{0.0075}=\frac{5000}{75}=\frac{200}{3}\approx 66.67{\text{ Nm}}^{-1}$

The spring constant is nearly $67{\text{ Nm}}^{-1}$.

Answer: $67$