The general displacement of a simple harmonic oscillator is x = A ( t). Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when t = T. The value of is:

Given: x = A ( t) for a simple harmonic oscillator. Find: The value of if the slope of the U-t curve is maximum at t = T. Potential energy is U = 12kx^2 Differentiating with respect to time, dUdt = 12k 2x dxdt Using x = A ( t), dUdt = kA^2 ( t) ( t) Now use 2 ( t) ( t) = (2 t) So, dUdt = kA^2 2 (2 t) For maximum slope, (2 t) = 1 Hence, 2 t = 2 So, t = 4 Using T = 2, t = T8 Therefore, = 8. Given: x = A ( t). Find: When the slope of the potential energy-time graph is maximum. Since U x^2 ^2( t) we have U 1 - (2 t)2 Therefore, dUdt (2 t) This is maximum when 2 t = 2 Thus, t = 4 = T8 So the required value is 8.

The general displacement of a simple harmonic oscillator is x = A… — Solution | JEE Physics

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The general displacement of a simple harmonic oscillator is $x = A \sin(\omega t)$ . Let $T$ be its time period. The slope of its potential energy ( $U$ ) - time ( $t$ ) curve will be maximum when $t = \frac{T}{\beta}$ . The value of $\beta$ is:

Answer

Correct answer:8

Step-by-step solution

Standard Method

Given: $x = A \sin(\omega t)$ for a simple harmonic oscillator.

Find: The value of $\beta$ if the slope of the $U$ - $t$ curve is maximum at $t = \frac{T}{\beta}$ .

Potential energy is

U = \frac{1}{2}kx^2

Differentiating with respect to time,

\frac{dU}{dt} = \frac{1}{2}k \cdot 2x \frac{dx}{dt}

Using $x = A\sin(\omega t)$ ,

\frac{dU}{dt} = kA^2\omega \sin(\omega t)\cos(\omega t)

Now use

2\sin(\omega t)\cos(\omega t) = \sin(2\omega t)

So,

\frac{dU}{dt} = \frac{kA^2\omega}{2}\sin(2\omega t)

For maximum slope,

\sin(2\omega t) = 1

Hence,

2\omega t = \frac{\pi}{2}

So,

t = \frac{\pi}{4\omega}

Using $T = \frac{2\pi}{\omega}$ ,

t = \frac{T}{8}

Therefore, $\beta = 8$ .

Using phase of energy directly

Given: $x = A\sin(\omega t)$ .

Find: When the slope of the potential energy-time graph is maximum.

Since

U \propto x^2 \propto \sin^2(\omega t)

we have

U \propto \frac{1 - \cos(2\omega t)}{2}

Therefore,

\frac{dU}{dt} \propto \sin(2\omega t)

This is maximum when

2\omega t = \frac{\pi}{2}

Thus,

t = \frac{\pi}{4\omega} = \frac{T}{8}

So the required value is $8$ .

Common mistakes

Taking potential energy as proportional to $x$ instead of $x^2$ . In SHM, $U = \frac{1}{2}kx^2$ , so the time dependence must come through $\sin^2(\omega t)$ , not $\sin(\omega t)$ .
Maximizing $\sin(\omega t)\cos(\omega t)$ incorrectly by setting either factor individually to $1$ . The product is best handled using $2\sin\theta\cos\theta = \sin 2\theta$ .
Using the condition for maximum potential energy instead of maximum slope of the $U$ - $t$ graph. The question asks for maximum value of $\frac{dU}{dt}$ , not maximum value of $U$ itself.

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