For a particle performing S.H.M., the total energy is ' n ' times the kinetic energy, when the displacement of a particle from mean position is 2 3 A, where A is the amplitude of S.H.M.

Question

For a particle performing S.H.M., the total energy is ' n ' times the kinetic energy, when the displacement of a particle from mean position is 2 3 ​ ​ A, where A is the amplitude of S.H.M. The value of ' n ' is

TestHub · Accepted Answer

Total energy of a particle in S.H.M. is given by $E = \frac{1}{2} m ω^{2} A^{2}$ .

Kinetic energy is given by $K = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$ .

Given displacement $x = \frac{3}{2} A$ .

Substituting the value of $x$ in the expression for kinetic energy:

$K = \frac{1}{2} m ω^{2} A^{2} - (\frac{3}{2} A)^{2}$

$K = \frac{1}{2} m ω^{2} (A^{2} - \frac{3}{4} A^{2})$

$K = \frac{1}{2} m ω^{2} (\frac{1}{4} A^{2})$

$K = \frac{1}{4} (\frac{1}{2} m ω^{2} A^{2})$

$K = \frac{1}{4} E$

$E = 4 K$

Comparing with $E = n K$ , we get $n = 4$ .

Answer: $4$

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