In S.H.M. the displacement of a particle at an instant is Y = Acos 30°, where A = 40 cm and kinetic energy is 200 J. If force constant is 1 × 10x N/m, then x will be (cos 30° = √(3)/2) (a) 4 (b) 3 (c) 2 (d) 1

Answer: (a). KE = 1/2k(A² - Y²) with Y² = 3/4A², so KE = kA²/8. 200 = (k(0.4)²)/(8) = 0.02k ⇒ k = 10⁴ N/m So x = 4; the answer is (a) 4.

In S.H.M. the displacement of a particle at an instant is Y = Acos… | MHT CET 2024 (9 May, Shift 1)

Q31MHT CET 2024 · 9 MAY · SHIFT 1Medium

Energy in SHM

In S.H.M. the displacement of a particle at an instant is $Y = A\cos 30^\circ$ , where $A = 40\text{ cm}$ and kinetic energy is 200 J. If force constant is $1 \times 10^{x}\text{ N/m}$ , then x will be $\left(\cos 30^\circ = \sqrt{3}/2\right)$

Show answer and solution

Answer: (a)

$KE = \frac{1}{2}k(A^2 - Y^2)$ with $Y^2 = \frac{3}{4}A^2$ , so $KE = \frac{kA^2}{8}$ .
$200 = \frac{k(0.4)^2}{8} = 0.02k \Rightarrow k = 10^4$ N/m
So $x = 4$ ; the answer is (a) 4.

Report a mistake

Energy in SHM question from MHT CET 2024 (9 May, Shift 1)

More questions on Energy in SHM