Question

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

• A. $1\times {10}^{2}N{m}^{-1}$
• B. $1.5\times {10}^{2}N{m}^{-1}$
• C. $0.0667\times {10}^{2}N{m}^{-1}$
• D. $3\times {10}^{2}N{m}^{-1}$

Answer 1

The correct option is B $1.5\times {10}^{2}N{m}^{-1}$

The Potential energy of the harmonic oscillator is given by $\text{PE}=\frac{1}{2}k{x}^2$

However, there can be an initial potential energy, which is a constant.

Thus, the generalized potential energy of a harmonic oscillator is $\text{PE}=V_0+\frac{1}{2}k{x}^2$

From the given figure, at $x=0$, $\text{PE}=0.01\text{J}$ and at $x=0.02\text{m}$, $\text{PE}=0.04\text{J}$

Substituting in the general form of Potential energy, we have $V_0=0.01\text{J}$

Thus, $\frac{1}{2}k(0.02{)}^2=0.03\Rightarrow k=150\text{N/m}$