JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Elasticity & SHM

At a given point of time the value of displacement of a simple harmonic oscillator is given as

y = A cos (3 0^{0})

. If amplitude is 40 cm and kinetic energy at that time is 200 J, the value of force constant is

1.0 \times 1 0^{x}

^{- 1}

. The value of

x

is ______.

Answer: 4

Solution

Given displacement

x = A cos (3 0^{\circ}) = 40 \times \frac{3}{2} = 203

= 0.2 3

m

Amplitude

A = 40

= 0.4

m

Kinetic energy in SHM:

K . E . = \frac{1}{2} k (A^{2} - x^{2})

200 = \frac{1}{2} k [(0.4)^{2} - (0.2 3)^{2}]

200 = \frac{1}{2} k [0.16 - 0.12]

200 = \frac{1}{2} k \times 0.04

k = \frac{400}{0.04} = 10000 = 1 0^{4} N/m

Therefore,

x = 4

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer