Question

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

• A. $1\times {10}^{2}N/m$
• B. $3\times {10}^{2}N/m$
• C. $0.667\times {10}^{2}N/m$
• D. $150N/m$

Answer 1

The correct option is D $150N/m$

From the diagram,
Total potential energy $U_T=0.04J$
Resisting potential energy $U_R=0.01J$
Amplitude $a=20mm$
Maximum Kintic energy,
$K_{max}=U_T-U_R\Rightarrow K_{max}=0.04J-0.01J\Rightarrow K_{max}=0.03J$
Kinatic energy can be expresed as,
$K_{max}=\frac{1}{2}m{\omega }^2{a}^2\Rightarrow K_{max}=\frac{1}{2}k{a}^2\therefore \text{spring costant}k=m{\omega }^2\Rightarrow 0.03=\frac{1}{2}k{a}^2\Rightarrow 0.03=\frac{1}{2}\times k\times (20\times {10}^{-3}{)}^2\Rightarrow k=\frac{0.06}{400\times {10}^{-6}}\Rightarrow k=150N/m$