Frequency of sonometer wire is given by, f=2L1μT
Therefore, f1=21μ6 f2=21μ54
Hence, f2f1=31
Also it is given that, f2−f1=12
So, f1=6Hz.
JEE Main 2024 — Physics Waves & Oscillations
A tuning fork resonates with a sonometer wire of length 1m stretched with a tension of 6N. When the tension in the wire is changed to 54N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is _______ Hz.
Held on 1 Feb 2024 · Verified 6 Jul 2026.
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
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.]
Work through every JEE Main Waves & Oscillations PYQ, year by year.