Q. No.The angular velocity and the amplitude of a simple pendulum is ω and a respectively. At a displacement X from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is
1, X2ω2(a2-X2ω2)
2. X2/(a2-x2)
3. (a2-X2ω2)/X2ω2
4. (a2-x2)/X2
1. f2
2. f
3. 2f
4. 4f
There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force ,F=-Kx where x is the displacement. The total energy of body depends upon -
1. K, x
2. K, a
3. K, a, x
4. K, a, v
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)
1. 18E
2. 14E
3. 12E
4. 23E
A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. Which of the following statements is true ?
1. P.E. is maximum when x = 0
2. K.E. is maximum when x = 0
3. T.E. is zero when x = 0
4. K.E. is maximum when x is maximum
A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration g4 , then the period of the pendulum will be
1. T
2. T4
3. 2T√5
4. 2T√5
The total energy of a particle, executing simple harmonic motion is
1. ∝ x
2. ∝ x2
3. Independent of x
4.∝ x1/2
The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle θ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals
1. √2gl(1-sinθ)
2. √2gl(1+cosθ)
3. √2gl(1-cosθ)
4. √2gl(1+sinθ)
A body is executing Simple Harmonic Motion. At a displacement x its potential energy is E1 and at a displacement y its potential energy is E2. The potential energy E at displacement (x+y) is
1. E=√E1+√E2
2. √E=√E1+√E2
3. E=E1+E2
4. None of these.
1. LT=constant
2. L2T=constant
3. LT2=constant
4. L2T2=constant
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