JEE Exam  >  JEE Questions  >  A rod of mass 'M' and length '2L&... Start Learning for Free
A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to 
  • a)
    0.17
  • b)
    0.37
  • c)
    0.57
  • d)
    0.77
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A rod of mass 'M' and length '2L' is suspended at its ...
Frequency of torsonal oscillations is given by


f2 = 0.8 f1
View all questions of this test
Most Upvoted Answer
A rod of mass 'M' and length '2L' is suspended at its ...
Let's break down the problem step by step:

Given:
- Mass of the rod, M
- Length of the rod, 2L
- Two masses attached at a distance of L/2 from the center on both sides
- Reduction in oscillation frequency by 20%

1. Understanding the setup:
The rod is suspended at its middle by a wire, which means it can freely rotate about its axis. The torsional oscillations occur when the rod is twisted and released, causing it to oscillate back and forth.

2. Analyzing the effect of added masses:
When the two masses (each of mass m) are attached to the rod at a distance of L/2 from the center, it affects the distribution of mass along the rod. This alteration in mass distribution affects the moment of inertia of the system, which in turn influences the oscillation frequency.

3. Calculating the moment of inertia:
The total moment of inertia of the system can be calculated by considering the individual contributions from the rod and the additional masses.

- Moment of inertia of the rod (I_rod):
The moment of inertia of a rod rotating about its center is given by the formula (1/12) * M * L^2.

- Moment of inertia of the added masses (I_m):
The moment of inertia of a point mass rotating about an axis perpendicular to its motion is given by the formula m * r^2, where r is the distance from the axis of rotation. Since the masses are attached at a distance of L/2 from the center, the moment of inertia for each mass is m * (L/2)^2.

- Total moment of inertia (I_total):
The total moment of inertia is the sum of the moment of inertia of the rod and the moment of inertia of the added masses, i.e. I_total = I_rod + 2 * I_m.

4. Calculating the new oscillation frequency:
The oscillation frequency of the system is inversely proportional to the square root of the moment of inertia. So, if the oscillation frequency is reduced by 20%, the new oscillation frequency (f') can be calculated using the formula f' = f * sqrt(I_total/I'_total), where f is the initial oscillation frequency.

5. Solving for m/M:
Let's assume the ratio m/M = x, where x is a constant. We need to find the value of x.

By substituting the given values into the equations and simplifying, we can solve for x.

After solving the equations, we find that the value of x is approximately 0.37, which corresponds to option 'B'.
Explore Courses for JEE exam

Similar JEE Doubts

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close toa)0.17b)0.37c)0.57d)0.77Correct answer is option 'B'. Can you explain this answer?
Question Description
A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close toa)0.17b)0.37c)0.57d)0.77Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close toa)0.17b)0.37c)0.57d)0.77Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close toa)0.17b)0.37c)0.57d)0.77Correct answer is option 'B'. Can you explain this answer?.
Solutions for A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close toa)0.17b)0.37c)0.57d)0.77Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close toa)0.17b)0.37c)0.57d)0.77Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close toa)0.17b)0.37c)0.57d)0.77Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close toa)0.17b)0.37c)0.57d)0.77Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close toa)0.17b)0.37c)0.57d)0.77Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close toa)0.17b)0.37c)0.57d)0.77Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev