# Chapter 14 – Oscillations Questions and Answers: NCERT Solutions for Class 11 Physics

Class 11 Physics NCERT book solutions for Chapter 14 - Oscillations Questions and Answers.

Education Blogs Chapter 14 – Oscillations Questions and Answers: NCERT Solutions for Class 11...

Class 11 Physics NCERT book solutions for Chapter 14 - Oscillations Questions and Answers.

(a) A swimmer completing one (return) trip from one bank of a river to the other and back.

(b) A freely suspended bar magnet displaced from its N-S direction and released.

(c) A hydrogen molecule rotating about its centre of mass.

(d) An arrow released from a bow.

(b) Since a freely suspended magnet if once displaced from N-S direction and released, it oscillates about this position, it is a periodic motion.

(c) The rotating motion of a hydrogen molecule about its centre of mass is periodic.

(d) Motion of an arrow released from a bow is non-periodic.

periodic but not simple harmonic motion?

(a) the rotations of earth about its axis.

(b) motion of an oscillating mercury column in a U-tube.

(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.

(d) general vibrations of a polyatomic molecule about its equilibrium position.

(b) It is S.H.M.

(c) It is S.H.M.

(d) General vibrations of a polyatomic molecule about its equilibrium position is periodic but non SHM. In fact, it is a result of superposition of SHMs executed by individual vibrations of atoms of the molecule.

(a) sin wt – cos wt (b) sin2wt (c) 3 cos -2 cos (π/4-2 wt) (d) cos wt + cos 3 wt + cos 5 wt

(e) exp (- w2t2) (f) 1 + wt + w2t2.

(f) 1 + wt + w2t2also represents non periodic motion.

(a) at the end A,

(b) at the end B,

(c) at the mid-point of AB going towards A,

(d) at 2 cm away from B going towards A,

(e) at 3 cm away from A going towards B, and (f) at 4 cm away from B going towards A.

(b) At the end B, i.e., second extreme position, velocity is zero whereas the acceleration and force are directed towards the point O and are negative.

(c) At the mid point O, while going towards A, velocity is negative and maximum. The acceleration and force both are zero.

(d) At 2 cm away from B, that is, at C and going towards A: v is negative; acceleration and F, being directed towards O, are also negative.

(e) At 3 cm away from A, that is, at D and going towards B: v is positive; acceleration and F, being directed towards O, are also positive.

(f) At a distance of 4 cm away from A and going towards A, velocity is directed along BA, therefore, it is positive. Since acceleration and force are directed towards OB, both of them are positive.

(a) a = 0.7x (b) a = – 200 x2

(c) a = – 10x (d) a = 100 x3

(x is in cm and t is in s)

(a) x = – 2 sin (3t + π /3)

(b) x = cos (π /6 – t)

(c) x = 3 sin (2πt + π /4)

(d) x = 2 cos π t.

(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released free, what is the period of oscillation in each case?

(d) During free fall of the cabin, the acceleration due to gravity is zero. Therefore, the frequency of oscillations is also zero i.e., the pendulum will not vibrate at all.

Consider the mercury contained in a vertical U-tube upto the level P and Q in its two limbs.

Let P = density of the mercury.

L = Total length of the mercury column in both the limbs.

A = internal cross-sectional area of U-tube. m = mass of mercury in U-tube = LAP.

Assume, the mercury be depressed in left limb to F by a small distance y, then it rises by the same amount in the right limb to position Q’.

.’. Difference in levels in the two limbs = P’ Q’ = 2y.

:. Volume of mercury contained in the column of length 2y = A X 2y

.•. m – A x 2y x ρ.

If W = weight of liquid contained in the column of length 2y.

Then W = mg = A x 2y x ρ x g

This weight produces the restoring force (F) which tends to bring back the mercury to its equilibrium position.

Increase the pressure on the ball by a little amount p, so that the ball is depressed to position D, where CD = y.

There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV = Ay

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