The displacement of a particle, executing simple harmonic motion with time period T, is expressed as where A is the amplitude. The maximum value of potential energy of this oscillator is found

Question

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 $t = \frac{T}{2 β} .$ The value of $β$ is ______. [2026]

Smart Achievers · Accepted Answer

(2)

Potential energy is maximum at extreme position. The particle starting at mean position reaches extreme position in time $\frac{T}{4}$ .

Solution

Q.
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 $t = \frac{T}{2 β} .$ The value of $β$ is ______. [2026]

Ans.
(2)

Potential energy is maximum at extreme position. The particle starting at mean position reaches extreme position in time $\frac{T}{4}$ .

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Solution

Q. The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t)=Asinωt where A is the amplitude. The maximum value of potential energy of this oscillator is found at t=T2β. The value of β is ______. [2026]

Ans. (2) Potential energy is maximum at extreme position. The particle starting at mean position reaches extreme position in time T4.

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Q.
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 $t = \frac{T}{2 β} .$ The value of $β$ is ______. [2026]

Ans.
(2)

Potential energy is maximum at extreme position. The particle starting at mean position reaches extreme position in time $\frac{T}{4}$ .