The velocity of a particle executing SHM varies with displacement $(\mathrm{x})$ as $4 \mathrm{v}^{2}=50-\mathrm{x}^{2}$. The time period of oscillations is $\frac{x}{7} s$. The value of $x$ is \ldots\ldots.. $\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Elasticity & SHM

The velocity of a particle executing SHM varies with displacement

(x)

4 v^{2} = 50 - x^{2}

. The time period of oscillations is

\frac{x}{7} s

. The value of

x

is \ldots\ldots..

(

Take

π = \frac{22}{7})

Answer: 88

Rewrite the given relation as

v^{2} = \frac{50 - x ^{2}}{4} = \frac{1}{4} (50 - x^{2})

and compare it with the standard form

v^{2} = ω^{2} (A^{2} - x^{2})

; the coefficient of

x^{2}

fixes

ω^{2} = \frac{1}{4}

, so

ω = \frac{1}{2}

rad s

^{- 1}

. The time period is

T = \frac{2 π}{ω} = 4 π = 4 \times \frac{22}{7} = \frac{88}{7}

s. Matching with

\frac{x}{7}

s, the value of

x

88

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Integer answer