The motion of a mass on a spring, with spring constant K is as shown in figure. The equation of motion is given by $x(t)=A \sin \omega t+B \cos \omega t$ with $\omega=\sqrt{\frac{K}{m}}$. Suppose that at time $t=0$, the position of mass is $x(0)$ and velocity $v(0)$, then its displacement can also be represented as $x(t)=C \cos (\omega t-\phi)$, where C and $\phi$ are :

(C) $C=\sqrt{\frac{v(0)^{2}}{\omega^{2}}+x(0)^{2}}, \phi=\tan ^{-1}\left(\frac{v(0)}{x(0) \omega}\right)$

(A) $C=\sqrt{\frac{v(0)^{2}}{\omega^{2}}+x(0)^{2}}, \phi=\tan ^{-1}\left(\frac{x(0) \omega}{v(0)}\right)$

(B) $C=\sqrt{\frac{2 v(0)^{2}}{\omega^{2}}+x(0)^{2}}, \phi=\tan ^{-1}\left(\frac{x(0) \omega}{2 v(0)}\right)$

(D) $C=\sqrt{\frac{2 v(0)^{2}}{\omega^{2}}+x(0)^{2}}, \phi=\tan ^{-1}\left(\frac{v(0)}{x(0) \omega}\right)$

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Elasticity & SHM

The motion of a mass on a spring, with spring constant K is as shown in figure. The equation of motion is given by

x (t) = A sin ω t + B cos ω t

with

ω = \frac{K}{m}

. Suppose that at time

t = 0

, the position of mass is

x (0)

and velocity

v (0)

, then its displacement can also be represented as

x (t) = C cos (ω t - ϕ)

, where C and

ϕ

are :

(A) $C = \frac{v ( 0 ) ^{2}}{ω ^{2}} + x (0)^{2}, ϕ = tan^{- 1} (\frac{x ( 0 ) ω}{v ( 0 )})$
(B) $C = \frac{2 v ( 0 ) ^{2}}{ω ^{2}} + x (0)^{2}, ϕ = tan^{- 1} (\frac{x ( 0 ) ω}{2 v ( 0 )})$
(C) $C = \frac{v ( 0 ) ^{2}}{ω ^{2}} + x (0)^{2}, ϕ = tan^{- 1} (\frac{v ( 0 )}{x ( 0 ) ω})$ ✓ Correct
(D) $C = \frac{2 v ( 0 ) ^{2}}{ω ^{2}} + x (0)^{2}, ϕ = tan^{- 1} (\frac{v ( 0 )}{x ( 0 ) ω})$

Answer: (C)

Expanding the second form,

C cos (ω t - ϕ) = C cos ϕ cos ω t + C sin ϕ sin ω t

, so comparison with

x (t) = A sin ω t + B cos ω t

gives

A = C sin ϕ

and

B = C cos ϕ

.
At

t = 0

the position is

x (0) = B = C cos ϕ

, and differentiating gives

v (0) = A ω = ω C sin ϕ

.
Squaring and adding the two relations,

x (0)^{2} + \frac{v ( 0 ) ^{2}}{ω ^{2}} = C^{2}

and dividing them,

tan ϕ = \frac{v ( 0 )}{ω x ( 0 )}

C = \frac{v ( 0 ) ^{2}}{ω ^{2}} + x (0)^{2}

and

ϕ = tan^{- 1} (\frac{v ( 0 )}{x ( 0 ) ω})

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option