x1=7sin5tx2=27sin(5t+3π)
From phasor,

∴ Amplitude of resultant SHM=7
ϕ=tan−17+27×2127×3/2=tan−12721=tan−123∴XR=7sin(5t+ϕ)aR=−7×25sin(5t+ϕ)∴amax=175 cm/sec=175×10−2 m/sec
JEE Main 2025 — Physics Waves & Oscillations
A particle is subjected two simple harmonic motions as :
x1=7sin5tcm
and x2=27sin(5t+3π)cm
where x is displacement and t is time in seconds. The maximum acceleration of the particle is x×10−2 ms−2. The value of x is :
Held on 2 Apr 2025 · Verified 6 Jul 2026.
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