SHM Formulae
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A particle is moving on x-axis has potential energy U=2−20x+5x2 J along x-axis. The particle is released at x=-3. The maximum value of x will be [x is in meter and U is in joule]
5 m
3 m
7 m
8 m
- Angular frequency of SHM will be 2 rad/sec
- Kinetic Energy at mean position =16 J
- Amplitude of SHM is 2 m
- None of these
A particle of mass 40 g executes a simple harmonic motion of amplitude 2.0 cm. If the time period is 0.20 s, find the total mechanical energy of the system.
7.9J
3.5 × 10−3J
7.9× 10−3J
35J
- 12mω2A2
- mω2A2
- 14mω2A2
- zero
Two blocks A and B each of mass m are connected by a massless spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with spring at its natural length as shown in the figure. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B and collides with A. Then,
The KE of the A-B system at maximum compression of the spring is zero
The KE of the AB system at maximum compression of the spring is mv22
The maximum compression of the spring is v√mk
The maximum compression of the spring is v√m2k
A particle is moving on x-axis has potential energy U=2−20x+5x2 J along x-axis. The particle is released at x=-3. The maximum value of x will be [x is in meter and U is in joule]
5 m
3 m
7 m
8 m
- 150 N/m
- 200 N/m
- 300 N/m
- 600 N/m