Oscillation and Simple Harmonic Motion | Physics JEE previous year questions with Solutions

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Q 1 :

A particle is subjected two simple harmonic motions as:

$x_{1} = \sqrt{7} \sin 5 t cm and x_{2} = 2 \sqrt{7} \sin (5 t + \frac{π}{3}) cm$

where x is displacement and t is time in seconds.

The maximum acceleration of the particle is $x \times 10^{- 2} {ms}^{- 2}$ . The value of x is: [2025]

175
$25 \sqrt{7}$
$5 \sqrt{7}$
125

(1)

Applied simple harmonic motions on the particle,

$x_{1} = \sqrt{7} \sin 5 t and x_{2} = 2 \sqrt{7} \sin (5 t + \frac{π}{3})$

From phasor,

Amplitude of resultant SHM,

$A = \sqrt{{(\sqrt{7})}^{2} + {(2 \sqrt{7})}^{2} + 2 \times 2 \times 7 \cos (\frac{π}{3})} = 7 cm$

The maximum acceleration of the particle

$a_{\max} = ω^{2} A = 25 \times 7 = 175 {cms}^{- 2} = 175 \times 10^{- 2} {ms}^{- 2}$

Q 2 :

For a periodic motion represented by the equation $Y = \sin ω t + \cos ω t$

The amplitude of the motion is [2023]

$0.5$
$\sqrt{2}$
$1$
$2$

(2)

$y = \sin ω t + \cos ω t$

$y = \sin ω t + \sin (ω t + \frac{π}{2})$

$Δ ϕ = \frac{π}{2}$

$A_{net} = \sqrt{1^{2} + 1^{2} + 2 \times 1 \times 1 \times \cos (Δ ϕ)}$

$A_{net} = \sqrt{2}$

Topic Question Set

Q 2 :

For a periodic motion represented by the equation $Y = \sin ω t + \cos ω t$

The amplitude of the motion is [2023]

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