The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t)=\mathrm{A} \sin \omega \mathrm{t}$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t=T / 2 \beta$. The value of $\beta$ is \_\_\_\_.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Elasticity & SHM

The displacement of a particle, executing simple harmonic motion with time period

T

, is expressed as

x (t) = A sin ω t

, where

A

is the amplitude. The maximum value of potential energy of this oscillator is found at

t = T /2 β

. The value of

β

is ____.

Answer: 2

For SHM

x = A sin ω t

, potential energy is maximum at the extreme positions where

x = \pm A

.

Starting from mean position (

x = 0

t = 0

), the particle first reaches an extreme position at

t = T /4

.

Given that this occurs at

t = T / (2 β)

\frac{T}{4} = \frac{T}{2 β}

Therefore:

β = 2

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Integer answer