NCERT Physics Class 11 Exemplar Ch 14 Oscillations Part 5 (For CBSE, ICSE, IAS, NET, NRA 2022)

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35. A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s – 1 and amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.

(a) Will there be any change in weight of the body, during the oscillation?

(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?

Ans: (a) Yes.

(b) Maximum weight


Minimum weight

Minimum weight

Maximum weight is at the topmost position,

Minimum weight is at the lowermost position.

36. A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand.

(a) What is the amplitude of oscillation?

(b) Find the frequency of oscillation?

Ans: (a) 2cm (b)

37. A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S. H. M. with a time period.

Where m is mass of the body and ρ is density of the liquid.

Ans: Let the log be pressed and let the vertical displacement at the equilibrium position be .

At equilibrium mg = Buoyant force

When it is displaced by a further displacement , the buoyant force is .

Net restoring force

. i.e.. proportional to .

38. One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

Ans: Consider the liquid in the length dx. it՚s mass is at a height x.

The PE of the left column

Similarly, P. E. of the right column

Where l is the length of the liquid in one arm of the tube.

Total P. E.

If the change in liquid level along the tube in left side in y, then length of the liquid in left side is and in the right side is .

Total P. E. 2 2 2 2

Change in

Change in K. E.

Change in total energy

Differentiating both sides w. r. t. time,

39. A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

Ans: Acceleration due to gravity at , where g is the acceleration at the surface.


Motion will be SHM with time period

40. A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground (Fig. 11) . The amplitude is . The string snaps at . Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume to be small so that and

Ans: Assume that when . Then,

Given a seconds pendulum

At time


Thus the linear velocity is perpendicular to the string.

The vertical component is

and the horizontal component is

At the time it snaps, the vertical height is

Let the time required for fall be t, then

(Notice g is also in the negative direction)

Or, Neglecting terms of order and heigher,


The distance travelled in the x direction is to the left of where it snapped.

To order of ,

At the time of snapping, the bob was

Distance from A.

Thus, the distance from A is

The Time Snapping