Given here, A=8cm and T=6s
The displacement equation in SHM, when amplitude is halved is Acos(T2πt)=2A
⇒T2πt=3π
Time taken by the particle is t= 6T=66=1sec.
A particle executes simple harmonic motion. Its amplitude is 8cm and time period is 6s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is _____ s
Held on 27 Jun 2022 · Verified 6 Jul 2026.
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The equation of a plane progressive wave is given by $y = 5\cos\pi\left(200t - \dfrac{x}{150}\right)$ where $x$ and $y$ are in cm and $t$ is in second. The velocity of the wave is _______ m/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.