A body executes simple harmonic motion with amplitude a. At what displacement from the equilibrium position is its energy half kinetic and half potential?

Question

A body executes simple harmonic motion with amplitude a. At what displacement from the equilibrium position is its energy half kinetic and half potential? (A)  (a/2)  (B)  (a/√2)  (C)  √2a  (D)  (a/3)

Tardigrade Admin · Accepted Answer

The kinetic energy (KE) of a body executing SHM with amplitude a undergoing displacement

y

is

K E = \frac{1}{2} m ω^{2} (a^{2} - y^{2}) \dots

(i)
where

ω

is angular velocity, and

m

the mass.
Also, potential energy (PE) is

PE = \frac{1}{2} m ω^{2} y^{2} \dots

(ii)
Given,

K E = PE

∴

\frac{1}{2} m ω^{2} (a^{2} - y^{2}) = \frac{1}{2} m ω^{2} y^{2}

\Rightarrow a^{2} - y^{2} = y^{2}

\Rightarrow y = \frac{a}{2}

Q. A body executes simple harmonic motion with amplitude $a$ . At what displacement from the equilibrium position is its energy half kinetic and half potential?

$\frac{a}{2}$

$\frac{a}{2}$

$2 a$

$\frac{a}{3}$

Q. A body executes simple harmonic motion with amplitude a. At what displacement from the equilibrium position is its energy half kinetic and half potential?

2a​

2​a​

2a​

3a​

The kinetic energy (KE) of a body executing SHM with amplitude a undergoing displacement y is KE=21​mω2(a2−y2)… (i) where ω is angular velocity, and m the mass. Also, potential energy (PE) is PE=21​mω2y2… (ii) Given, KE=PE ∴ 21​mω2(a2−y2)=21​mω2y2 ⇒a2−y2=y2 ⇒y=2​a​

Q. A body executes simple harmonic motion with amplitude $a$ . At what displacement from the equilibrium position is its energy half kinetic and half potential?

$\frac{a}{2}$

$\frac{a}{2}$

$2 a$

$\frac{a}{3}$