Question

The variation of potential energy of a harmonic oscillator is as shown in figure. The spring constant $k$ is

• A. $1\times {10}^2\text{N/m}$
• B. $1.5\times {10}^2\text{N/m}$
• C. $2\times {10}^2\text{N/m}$
• D. $3\times {10}^2\text{N/m}$

Answer 1

The correct option is B $1.5\times {10}^2\text{N/m}$

Potential energy of a particle executing simple harmonic motion is given by
$U-U_{mean}=\frac{1}{2}k{x}^2$
From the data given in the diagram, we can say that,
At $x=0.02\text{m},U=0.04\text{J}$
At $x=0,U=0.01\text{J}$
Using this we get,
$0.04-0.01=\frac{1}{2}k(0.02{)}^2\Rightarrow k=\frac{2\times 0.03}{(0.02{)}^2}=150\text{N/m}$
Thus, option (b) is the correct answer.