SHMEasy

Question

A linear harmonic oscillator of force constant 2 × 106 N/m and amplitude 0.01 m has a total mechanical energy of 160 J. Its

Options

A.maximum potential energy is 100 J
B.maximum kinetic energy is 100 J
C.maximum potential energy is 160 J
D.maximum potential energy is zero

Solution

kA2 = × 2 × 106 × (10-2)2 = 100J
This is basically the energy of oscillation of the particle. K, U and E at mean position (x = 0) and extreme position (x = A) are shown in figure.

∴ Correct option are (b) and (c).

Create a free account to view solution

View Solution Free
Topic: SHM·Practice all SHM questions

More SHM Questions

In the above question, the value of maximum velocity of the particle will be :...A body is executing simple harmonic motion with an angular frequency 2 rad/s. The velocity of the body at 20 mm displace...The equation of a simple harmonic motion is x = 0.34cos(3000t + 0.74). Where x and t are in mm and sec. respectively. Th...The average speed of the bob of a simple pendulum oscillating with a small amplitude A. and time period T is...The quantities which are always Zero for a particle performing linear SHM :...