At a given point of time the value of displacement of a simple harmonic oscillator is given as y = Acos (30°). If amplitude is 40 cm and kinetic energy at that time is 200 J, the value of force constant is 1.0 × 10x N m⁻¹. The value of x is ____________. Numerical answer type. Work it out, then open the answer.

Answer: 4. y² = A²cos ² 30° = 3/4A² KE = 1/2k(A² - y²) = kA²/8 200 = (k(0.4)²)/(8) = 0.02k ⇒ k = 1.0 × 10⁴ N m ⁻¹ So x = 4.

At a given point of time the value of displacement of a simple… | JEE Main 2023 (13 Apr, Shift 1)

Q45JEE Main 2023 · 13 APR · SHIFT 1Medium

Energy 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 $1.0 \times 10^{x}\text{ N m}^{-1}$ . The value of $x$ is ____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

$y^2 = A^2\cos^2 30^\circ = \frac{3}{4}A^2$
$KE = \frac{1}{2}k(A^2 - y^2) = \frac{kA^2}{8}$
$200 = \frac{k(0.4)^2}{8} = 0.02k \Rightarrow k = 1.0 \times 10^4$ N m $^{-1}$
So $x = 4$ .

Report a mistake

Energy in SHM question from JEE Main 2023 (13 Apr, Shift 1)

More questions on Energy in SHM