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.
175
257
57
125
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
In an open organ pipe $\nu_{3}$ and $\nu_{6}$ are $3^{\text {rd }}$ and $6^{\text {th }}$ harmonic frequencies, respectively. If $\nu_{6}-\nu_{3}=2200 \mathrm{~Hz}$ then length of the pipe is $\_\_\_\_$ mm . (Take velocity of sound in air is $330 \mathrm{~m} / \mathrm{s}$.)
A transverse wave on a string is described by $y = 3\sin(36t + 0.018x + \pi/4)$, where $x, y$ are in cm and $t$ in seconds. The least distance between the two successive crests in the wave is _____ cm. (Nearest integer) ($\pi = 3.14$)
Two waves of same frequency and amplitude travel in opposite directions. The resulting pattern is:
A spring-mass system oscillates with angular frequency ω. If the mass is doubled and the spring constant is halved, the new angular frequency is:
The time period of a simple harmonic oscillator is $T=2 \pi \sqrt{\frac{k}{m}}$. Measured value of mass $(m)$ of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant $(k)$ is $\_\_\_\_$ \%.
Work through every JEE Main Waves & Oscillations PYQ, year by year.