
L=2⋋⇒⋋=2L=1.2m
v=ρy=5.86×103ms−1
∴Fundamentalfrequency=⋋v=4.88kHz
≈5kHz
JEE Main 2018 — Physics Waves & Oscillations
A granite rod of 60cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7×103kgm−3 and its Young's modulus is 9.27×1010Pa. What will be the fundamental frequency of the longitudinal vibrations?
Held on 8 Apr 2018 · Verified 6 Jul 2026.
7.5kHz
5kHz
2.5kHz
10kHz
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.