NVAEasyJEE 2023Energy in SHM

JEE Physics 2023 Question with Solution

At a given point of time the value of displacement of a simple harmonic oscillator is given as y=Acos(30)y = A \cos (30^\circ). If amplitude is 40cm40 \, \text{cm} and kinetic energy at that time is 200J200 \, \text{J}, the value of force constant is 1.0×10xNm11.0 \times 10^x \, \text{N} \text{m}^{-1}. The value of xx is _____

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given: y=Acos30y = A \cos 30^\circ, amplitude A=40cm=0.4mA = 40 \, \text{cm} = 0.4 \, \text{m}, and kinetic energy KE=200JKE = 200 \, \text{J}.

Find: The value of xx in k=1.0×10xNm1k = 1.0 \times 10^x \, \text{N} \text{m}^{-1}.

For SHM,

KE=12k(A2y2)KE = \frac{1}{2} k (A^2 - y^2)

First, calculate displacement:

y=Acos30=0.4×32=0.3464my = A \cos 30^\circ = 0.4 \times \frac{\sqrt{3}}{2} = 0.3464 \, \text{m}

Now substitute in the kinetic energy expression:

200=12k(0.420.34642)200 = \frac{1}{2} k \left(0.4^2 - 0.3464^2\right)

Using

0.42=0.16,0.34642=0.1190.4^2 = 0.16, \quad 0.3464^2 = 0.119

we get

200=12k(0.160.119)200 = \frac{1}{2} k (0.16 - 0.119) 200=12k×0.041200 = \frac{1}{2} k \times 0.041

So,

k=200×20.041=4000.0419756.1Nm1k = \frac{200 \times 2}{0.041} = \frac{400}{0.041} \approx 9756.1 \, \text{N} \text{m}^{-1}

Thus,

k1.0×104Nm1k \approx 1.0 \times 10^4 \, \text{N} \text{m}^{-1}

Therefore, the value of xx is 44.

Common mistakes

  • Using KE=12kA2KE = \frac{1}{2}kA^2 directly is incorrect because that is the total energy, not the kinetic energy at displacement yy. Instead, use KE=12k(A2y2)KE = \frac{1}{2}k(A^2-y^2).

  • Not converting amplitude from 40cm40 \, \text{cm} to 0.4m0.4 \, \text{m} gives a wrong value of kk. Always convert to SI units before substitution.

  • Taking displacement as Asin30A \sin 30^\circ instead of Acos30A \cos 30^\circ is wrong because the question explicitly gives y=Acos30y = A \cos 30^\circ. Use the stated relation exactly.

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