JEE Main 2026 — Physics Waves & Oscillations
The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t)=Asinωt, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t=T/2β. The value of β is ____.
Held on 28 Jan 2026 · Verified 6 Jul 2026.
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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 $\_\_\_\_$ \%.
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:
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 simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is $\_\_\_\_$ cm. [Assume that the mass of the pendulum remains same.]
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