SHM Formulae
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Infinite number of bodies, each of mass 2kg are situated on X-axis at distance 1m, 2m, 4m, 8m and so on, respectively from the origin. The resulting gravitational potential due to this system at the origin will be:
- −4G
- −G
- −8G3
- −4G3
- l0+ilB−mgK
- l0+ilB−mg2K
- l0+mg−ilB2K
- l0+mg−ilBK
The total energy of the body executing SHM is E. Then the kinetic energy when the displacement is half of the amplitude is?
An object of mass 0.2 kg executes simple harmonic motion along X-axis with frequency of 25/ π Hz. At the position x = 0.04 m, the object has kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation in meter is equal to
0.05 m
0.06 m
0.01 m
0.002 m
A particle of mass 10g is executing simple harmonic motion with an amplitude of 0.5m and periodic time of π5s. The maximum value of the force acting on the particle is
- 25 N
- 5 N
- 2.5N
- 0.5 N
- The amplitude of oscillation in the first case changes by a factor of √Mm+M, whereas in the second case it remains unchanged
- The final time period of oscillation in both the cases is same
- The total energy decreases in both the cases
- The instantaneous speed at x0 of the combined masses decreases in both the cases
- 23E
- 18E
- 14E
- 12E
- 12mω2A2
- mω2A2
- 14mω2A2
- zero
The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2 cm, 1 m s−1 and 10 m s−2 at a certain instant. Find the amplitude and the time period of the motion.
What is the quartile skewness coefficient?
- n2
- 1n2
- n
- 1n
- V08
- V02
- V03
- V04
- 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
Consider the situation shown in figure (12-E10). Show that if the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. Calculate the time period.
- Amplitude of oscillation remains unchanged
- Time period of oscillation remains unchanged
- The total mechanical energy of the system does not change
- The maximum speed of the oscillating object changes
- √gl
- √gl2
- √2gl
- √3gl
A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal ?
- 0.5 J
- 0.6 J
- 0.3 J
- 0.8 J
- 4 cm
- 2 cm
- 2√2 cm
- √2 cm
- At 0.05 m from the origin is 50 m/s2
- At 0.05 m from the mean position is 100 m/s2
- At 0.05 m from the origin is 150 m/s2
- At 0.05 m from the mean position is 200 m/s2
Two identical blocks and each of mass resting on the smooth horizontal floor are connected by a light spring of natural length and spring constant . A third block of mass moving with a speed along the line joining and collides elastically with . The maximum compression in the spring is:
- the extreme positions and mean positions
- the mean position and extreme positions
- the mean position alternatively
- the extreme position alternatively
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.
35J
7.9J
3.5 x 10-3J
7.9 x 10-3J
- 150 N/m
- 200 N/m
- 300 N/m
- 600 N/m
A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2A3 from equilibrium position. The new amplitude of the motion is:
- 0.6A cm
- 0.5A cm
- 0.7A cm
- 0.9A cm
- not a simple harmonic
- simple harmonic with amplitude √a2+b2
- simple harmonic with amplitude ab
- simple harmonic with amplitude (a+b)2
- 2 : 1
- 4 : 1
- 1 : 1
- 1 : 2