JEE Main 2025 Apr 2 Shift 1, Physics Q9: Simple Harmonic Motion And Oscillation

JEE Main2025Apr 2, Shift 1Physics

A particle is subjected two simple harmonic motions as :

$x_{1} = 7 sin 5 t$ cm

and $x_{2} = 27 sin (5 t + \frac{π}{3})$ cm

where $x$ is displacement and $t$ is time in seconds. The maximum acceleration of the particle is $x \times 1 0^{- 2} ms^{- 2}$ . The value of $x$ is :

A
175
B
$257$
C
$57$
D
125

Solution

Two SHMs of the same angular frequency superpose into a single SHM with amplitude given by the phasor sum:

$A = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos ϕ$

With $A_{1} = 7$ , $A_{2} = 27$ , $ϕ = π /3$ :

$A^{2} = 7 + 28 + 2 \cdot 7 \cdot 27 \cdot \frac{1}{2} = 7 + 28 + 14 = 49$

$A = 7 cm$

Maximum acceleration of SHM:

$a_{m a x} = ω^{2} A = (5)^{2} \times 7 = 175 cm/s^{2} = 175 \times 1 0^{- 2} m/s^{2}$

$∴$ $x = 175$ . Option (1) is correct.

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JEE Main 2025 Apr 2 Shift 1, Physics Q9: Simple Harmonic Motion And Oscillation

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