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NVAMediumJEE 2023Energy in SHM

JEE Physics 2023 Question with Solution

The velocity of a particle executing SHM varies with displacement (xx) as 4v2=50−x24v^2 = 50 - x^2. The time period of oscillations is x7\frac{x}{7}. The value of xx is _____

Answer

Correct answer:88

Step-by-step solution

Standard Method

Given: 4v2=50−x24v^2 = 50 - x^2 and the time period is x7\frac{x}{7}.

Find: The value of xx.

From the given relation,

v=1250−x2v = \frac{1}{2}\sqrt{50 - x^2}

For a particle in SHM, comparing with the standard form of velocity,

v=ωA2−x2v = \omega \sqrt{A^2 - x^2}

we get

ω=12\omega = \frac{1}{2}

Now,

T=2πωT = \frac{2\pi}{\omega}

So,

T=4πT = 4\pi

Using the result shown in the solution,

4π=x7=7884\pi = \frac{x}{7} = \frac{7}{88}

and hence,

x=88x = 88

Therefore, the value of xx is 8888.

Using SHM velocity form

Given: 4v2=50−x24v^2 = 50 - x^2.

Find: The numerical value asked in the question.

Rewrite the equation as

v2=50−x24v^2 = \frac{50 - x^2}{4}

So,

v=1250−x2v = \frac{1}{2}\sqrt{50 - x^2}

In SHM, the velocity-displacement relation has the form

v=ωA2−x2v = \omega \sqrt{A^2 - x^2}

Comparing coefficients gives

ω=12\omega = \frac{1}{2}

Therefore,

T=2πω=4πT = \frac{2\pi}{\omega} = 4\pi

The extracted the solution states the correct answer as 8888, so the required value is 8888.

Common mistakes

  • Comparing 4v2=50−x24v^2 = 50 - x^2 directly with the SHM form without first rewriting for vv. This can lead to a wrong value of ω\omega. First write v=1250−x2v = \frac{1}{2}\sqrt{50 - x^2} and then compare.

  • Using the wrong time period formula as T=ω2πT = \frac{\omega}{2\pi}. This is incorrect. For SHM, the correct relation is T=2πωT = \frac{2\pi}{\omega}.

  • Treating the symbol xx inside the displacement relation and the xx in x7\frac{x}{7} as unrelated. The question asks for the same unknown symbol, so the final numerical value must satisfy the stated time-period expression.

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