4λ1=11+e⇒512×4v=11+e...(1)
4λ2=27+e⇒256×4v=27+e...(2)
Subtracting the equations,
256×4v×21=16cm=0.16m
⇒v=0.16×2×4×256
=328ms−1
JEE Main 2019 — Physics Waves & Oscillations
A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in air. A tuning fork of frequency 512Hz produces first resonance when the tube is filled with water to a mark 11cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256Hz which produces first resonance when water reaches a mark 27cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to
Held on 12 Jan 2019 · Verified 6 Jul 2026.
335ms−1
341ms−1
322ms−1
328ms−1
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.