The veriation of potential energy of harmonic escillator is as shown in figure. The spring constant is

Options

A. `1 xx 10^(2)N//m`
B. `150N//m`
C. `0.667 xx 10^(2)N//m`
D. `3 xx 10^(2)N//m`

Answer

B

Explanation

Total potential energy `= 0.04 J`
Resting potential energy `= 0.01 J`
Maximum kinetic energy `= (0.04 - 0.01)`
`= 0.03 J = (1)/(2)m omega^(2) a^(2) = (1)/(2) ka^(2)`
`0.03 = (1)/(2) xx 4 xx ((20)/(1000))^(2)`
`k = 0.06 xx 2500 Nm = 150 N//m`