The displacement of a simple harmonic motion | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given,amplitude of oscillation,$a = 6 	ext{ cm}$.Let the displacement be $x 	ext{ cm}$.When the kinetic energy $(K)$ is equal to the potential e

Vedclass · Accepted Answer

$3 \sqrt{2} 	ext{ cm}$

Vedclass · Answer

(C) Given,amplitude of oscillation,$a = 6 	ext{ cm}$.Let the displacement be $x 	ext{ cm}$.When the kinetic energy $(K)$ is equal to the potential energy $(U)$,we have $K = U$.The kinetic energy of a simple harmonic oscillator is given by $K = \frac{1}{2} m \omega^2 (a^2 - x^2)$.The potential energy is given by $U = \frac{1}{2} m \omega^2 x^2$.Equating the two:$\frac{1}{2} m \omega^2 (a^2 - x^2) = \frac{1}{2} m \omega^2 x^2$$a^2 - x^2 = x^2$$2x^2 = a^2$$x^2 = \frac{a^2}{2}$$x = \frac{a}{\sqrt{2}}$Substituting $a = 6 	ext{ cm}$:$x = \frac{6}{\sqrt{2}} = 3 \sqrt{2} 	ext{ cm}$.

The displacement of a simple harmonic motion of amplitude $6 \text{ cm}$ when its kinetic energy is equal to its potential energy is:

Similar Questions

When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation,the displacement of the particle from the equilibrium position in terms of its amplitude $a$ is

$A$ particle undergoing simple harmonic motion has an amplitude of $10 \ cm$. When the particle is at a displacement of $6 \ cm$ from the centre,then the ratio of its kinetic energy to potential energy is

$A$ block is executing $SHM.$ Let the time period of variation of velocity be $T_1$ and time period of variation of kinetic energy be $T_2.$ The relation between $T_1$ and $T_2$ is

$A$ body is executing simple harmonic motion with frequency $n$,the frequency of its potential energy is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module