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A spring has a certain mass suspended from it and its period for vertical oscillation is T_{1}. The spring is now cut into two equal halves and the same mass is suspended from one of the half. The period of vertical oscillation is now T_{2}. The ratio of is :
When spring is cut in equal part, then force constant is
The correct answer is: 0.707
For a simple harmonic vibrator of frequency n, the frequency of kinetic energy changing completely to potential energy is αn. Find the value of α.
2n
The correct answer is: 2
The total energy of the body executing S.H.M is E. Then the kinetic energy when the displacement is half of the amplitude is αE Find the value of α.
Kinetic energy
Kinetic energy
The correct answer is: 0.75
The amplitude of a damped harmonic oscillator become halved in 1 minute. After three minutes the amplitude will become 1/x of initial amplitude where x is :
The variation in amplitude of a damped harmonic oscillation with time is given by
initial amplitude,
damped factor
It is given that after 1 minutes
The correct answer is: 8
One body of mass m is suspended from three springs as shown in figure each spring has spring constant k. If mass m is displaced slightly then time period of oscillation is Find the value of α.
Using series and parallel combination of spring
The correct answer is: 1.5
A particle is executing SHM with an amplitude 4 cm. The displacement (in cm) at which its energy it half kinetic and half potential is :
Potential energy = Kinetic energy
The correct answer is: 2.828
A particle executing S.H.M of amplitude 4 cm and T = 4 sec. The time taken (in sec) by it to move from positive extreme position to half the amplitude is :
The correct answer is: 0.667
The potential energy of a particle executing S.H.M. is 2.5 J when its displacement is half of amplitude. The total energy (in Joule) of the particle is :
Total energy
Potential energy
So total energy
The correct answer is: 10
If the time period of oscillation of mass M suspended from a spring is one second, then the time period (in sec) of 4M will be :
The correct answer is: 2
A linear harmonic oscillation of force constant 2 x 10^{6} Nlm and amplitude 0.01 m has a total mechanical energy of 160 joules. Its maximum K.E. (in Joule) will be.
Maximum kinetic energy is 100 J
Maximum kinetic energy
The correct answer is: 100
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