99 Percentile Qs Bank for JEE MainPhysicsOscillations
The variation of potential energy of a harmonic oscillator is as shown in the figure. Then, find the spring constant.
Options
- A(1 10^2 Nm ⁻¹ )
- B(150 Nm ⁻¹ )
- C(0.667 10^2 Nm ⁻¹ )
- D(3 10^2 Nm ⁻¹ )
Correct answer
B. (150 Nm ⁻¹ )
Step-by-step solution
According to figure, when (y=20 ~mm ) (=2 10⁻² ~m ) then (U_ =0.04 ~J =4 10⁻² ~J ) When (y=0 ) then (U_ =0.01 ~J =1 10⁻² ~J ) ( ) The change in potential energy, ( array ll & U_ -U_ = 1 2 K y^2 & 4 10⁻²-1 10⁻²= 1 2 K (2 10⁻² )^2 & 3 10⁻²=K 2 10⁻⁴ & K= 3 10⁻² 2 10⁻⁴ =1.5 10^2=150 Nm ⁻¹ array )