T=2πKM 
35 T=2πKM+m
Dividing equation (ii) by equation (i), 35=MM+m. Squaring both the sides 925=MM+m=1+Mm⇒Mm=925−1=916
JEE Main 2003 — Physics Waves & Oscillations
A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 35 T. Then the ratio of Mm is
Held on 30 Apr 2003 · Verified 6 Jul 2026.
53
925
916
35
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.