Question

calculate the displacement to amplitude ratio of S.H.M. when kinetic energy is 90% of total energy. ​

Answer 1

Answer:

Let us assume

The mass of the particle/body executing simple harmonic motion be “m”

Amplitude be “a”

Angular Velocity be “ω”

The total energy of the particle to be 100%.

∴ The total energy = ½ * m * a² * ω² = 100% …. (i)

We are given that, K.E. = 90% of the total energy. So, let’s say the displacement of the particle, in this case, be denoted as “d”.  

Also, the potential energy = total energy – 90% of K.E. = 10% of K.E.  

∴ Potential energy = ½ * m * ω² * d² = 10% …… (ii)

Dividing equation (ii) by (i), we get

P.E. / total energy = 10% / 100%

Or, [½ * m * ω² * d²] / [½ * m * a² * ω²] = 10/100

Or, d² / a² = 1/10

Or, d/a = $\sqrt{0.1}$ = 0.316

Thus, the value of the displacement to amplitude ratio of S.H.M is 0.316.