Explanation
Motion of load attached to the spring is in SHM. The motion of liquid in U-tube is also in SHM.
It is very clear that the time period of the block is given by T = Time taken (t1) by the block to travel 30.0 cm towards right + time (t2) for which block remains in contact with the spring k2 + time taken (t3) by the block to travel 60 cm towards left + time (t4) for which the block remains in contact with spring k1 + time (t5) to travel 30 cm back,
i.e. T = t1 + t2 + t3 + t4 + t5
Force (F) is defined as the negative value of potential energy gradient i.e. F = – dU/dx.
The maximum value of K.E. = K0 maximum value of P.E.
The frequency of K.E. is twice that of a particle executive S.H.M., hence graph (a) is correct.
When the bob oscillates, the restoring force mg sin θ is always ⊥r to the length of string. So Workdone by string in one oscillation is zero.
Since K = K0 cos2ωt, therefore, maximum value of K.E. = K0 = Max. value of P.E. at the extreme position of the body in S.H.M.
At B, the velocity of maximum. Taking vertical downward motion of body from A to B, we have, u = 0, a = g, s = H and v = ?
The acceleration of the vehicle down the plane = g sinα. The reaction force acting on the bob of pendulum gives it an acceleration a (= g sinα) up the plane. This acceleration has two rectangular components, ax = a cosα = g sinα cosα and ay = a sinα =g sin2α as shown in Fig. The effective acceleration due to gravity acting on the bob is given by
Maximum KE in SHM is at mean position and maximum PE is at extreme position.
Here pendula A and C will be in resonance, hence will vibrate with maximum amplitude.
In SHM, the velocity of the particle is maximum at the mean position. Hence K.E. is maximum, when x = 0.
The displacement in SHM is given by g = r sin ωt and force
F = – mω2rsinωt = mω2rsin (ωt + π).
Thus force-time graph will be as shown in fig. (4).
On plotting a graph between P.E. and time t, we shall get a curve II and graph between PE and x we shall get curve III.
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