At a given point of time the value of displacement of a simple harmonic oscillator is given as y = A (30^). If amplitude is 40 cm and kinetic energy at that time is 200 J, the value of force constant…

At a given point of time the value of displacement of a simple ha… - Solution | JEE Physics

NVAEasyJEE 2023Energy in SHM

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

Answer

Correct answer:4

Step-by-step solution

Standard Method

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

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

For SHM,

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

First, calculate displacement:

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

Now substitute in the kinetic energy expression:

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

Using

0.4^2 = 0.16, \quad 0.3464^2 = 0.119

we get

200 = \frac{1}{2} k (0.16 - 0.119)

200 = \frac{1}{2} k \times 0.041

So,

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

Thus,

k \approx 1.0 \times 10^4 \, \text{N} \text{m}^{-1}

Therefore, the value of $x$ is $4$ .

Common mistakes

Using $KE = \frac{1}{2}kA^2$ directly is incorrect because that is the total energy, not the kinetic energy at displacement $y$ . Instead, use $KE = \frac{1}{2}k(A^2-y^2)$ .
Not converting amplitude from $40 \, \text{cm}$ to $0.4 \, \text{m}$ gives a wrong value of $k$ . Always convert to SI units before substitution.
Taking displacement as $A \sin 30^\circ$ instead of $A \cos 30^\circ$ is wrong because the question explicitly gives $y = A \cos 30^\circ$ . Use the stated relation exactly.

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