NCERT Solution (Class 11) > Physics Chapter 14 Oscillations
NCERT Solution Class 11 Physics Chapter 14 Oscillations: National Council of Educational Research and Training Class 11 Physics Chapter 14 Solution – Oscillations. Free PDF Download facility available at our website.
Board |
NCERT |
Class |
11 |
Subject |
Physics |
Chapter |
14 |
Chapter Name |
Oscillations |
Topic |
Exercise Solution |
Oscillations Chapter all Questions and Numericals Solution
Chapter 14- Oscillations
14.1) Which of the following examples represent periodic motion? (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.
Answer-
(a)The swimmer’s motion is not periodic. Swimmer may not take same time in upstream and downstream.
(b) When magnet is freely-suspended and released, it will oscillate about its position with a definite period of time hence motion is periodic in nature.
(c)When a hydrogen molecule rotates about its centre of mass, after an equal interval of time it comes to the same position again and again. Hence it represents periodic motion.
(d)An arrow released from a bow does not come back hence; this motion is not a periodic.
14.2) Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation 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 lower most point. (d) general vibrations of a polyatomic molecule about its equilibrium position.
Answer-
Periodic motion is the motion which repeats itself after equal interval of time while simple harmonic motion is the motion in which body performs to and fro motion about mean position in force is always directed towards mean position.
- During rotation of earth about its axis, earth repeats its motion after equal interval of time(1 day). Hence, it is a periodic motion. However, this motion is not simple harmonic as it is not a to and fro motion about its axis.
- An oscillating mercury column in a U-tube is simple harmonic. the mercury moves to and fro on the same path, about the fixed position.
- The ball bearing inside a smooth curved bowl moves to and fro about the lowermost point.And motion repeats itself after equal interval of time. Hence, its motion is periodic as well as simple harmonic.
- A polyatomic molecule has many natural frequencies of oscillation and vibrates with same time period hence it is periodic motion.
14.3) Fig. 14.23 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)
Answer-
(a) There is no repetition of motion in this case hence It is not a periodic motion.
(b) Here motion is periodic motion as it repeats itself after 2 s. Hence, it is a periodic motion with a period of 2 s.
(c) Motion of particle is random in nature hence It is not a periodic motion.
(d) Given wave are sine wave and the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion with a period of 2 s.
14.4) Which of the following functions of time represent
(a) simple harmonic,
(b) periodic but not simple harmonic, and
(c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant):
(a) sin ωt – cos ωt
(b) sin3 ωt
(c) 3 cos (π/4 – 2ωt)
(d) cos ωt + cos 3ωt + cos 5ωt
(e) exp (–ω2 t2 ) (f) 1 + ωt + ω2 t2
Answer-
Hence given function represents the simple harmonic motion. With period 2π/ω .
14.5 A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (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.
Answer-
(a)The given situation is shown in the following figure. Points A and B are the two end points, with AB = 10 cm. O is the midpoint of the path. A particle is in linear simple harmonic motion between the end points .
At the extreme point A, the particle is at rest Hence, its velocity is zero at this point. In SHM the force and acceleration is always towards mean position that is towards the point 0 hence acceleration is directed along OA and towards point O and same for the force.
(b) At the extreme point B, the particle is at rest Hence, its velocity is zero at this point. In SHM the force and acceleration is always towards mean position that is towards the point 0 hence acceleration is directed along OB and towards point O and same for the force which is directed leftwards.
(c) when particle At the middle point 0 and going towards point A , at middle velocity is maximum and negative as going towards left direction. hence acceleration and force both are zero at mean position.
(d) when particle at 2 cm away from B going towards Aas the particle going leftward hence velocity is negative and In SHM the force and acceleration is always towards mean position that is towards the point 0 hence acceleration is directed leftward and hence negative and same for the force which is directed leftwards hence negative.
14.6) Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? (a) a = 0.7x (b) a = –200x 2 (c) a = –10x (d) a = 100x 3 202
ANSWER-
A motion represents simple harmonic motion
F = – kx
∴ ma = – kx
a = (-k)/m x X
On comparing, Option (c) a = –10x represents the simple harmonic motion.
14.7) The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + φ ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π rad/s. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Answer:
Given that at t = 0,
x = 1 cm
v = ω cm/s
ω =π rad/s
The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + φ )
∴ 1 = A cos (0 + φ)
∴1 = A cos (φ) ……….(1)
In case of SHM, Velocity is given by
V = -Aω Sin (ωt + φ )
∴ ω =-Aω Sin (ωt + φ )
∴1 =-A Sin (0 + φ )
∴-1 =A Sin ( φ ) ……………(2)
Squaring and adding equation 1 and 2 we get,
∴A2 Sin2 (φ) +A2 cos2 (φ) = 1 + 1
∴A2 = 2
∴A = √2 cm
To find the phase angle divide equation 2 by 1,
tan (φ) = -1
∴(φ) = 3π/4 .
If instead of the cosine function, we choose the sine function to describe the SHM:
x = B sin (ωt + α)
Given that at t = 0,
x = 1 cm
v = ω cm/s
ω = π rad/s
The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = B sin (ωt + α)
∴1 = B sin (0 + α)
∴1 = B sin (α)……….(1)
In case of SHM, Velocity is given by
V =Bωcos (ωt + α)
∴ω =Bω cos (0 + α)
∴1 =B cos( α)
∴1 =Bcos( α)……………(2)
Squaring and adding equation 1 and 2 we get,
∴B2 sin2 (α) + B2cos2 (α) = 1 + 1
∴ B2 = 2
∴ B = cm
To find the phase angle divide equation 2 by 1,
tan (α) = 1
∴(α) =p/4 .
14.8)
14.9)
14.11) Figures 14.25 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure. Fig. 14.25 Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
ANSWER-
From figure a; TIME PERIOD = 2 second and amplitude (A)= 3cm
At time, t = 0, the radius vector OP makes an angle of 900 with x axis. Hence phase angle is 900.
Therefore, the equation of simple harmonic motion is given by,
14.12) Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t =0) position of the particle, the radius of the circle and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (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 πt
ANSWER-
(a) x = –2 sin (3t + π/3)
Standard equation of simple harmonic motion is given by,
14.13) Figure 14.26 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.26 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in n Fig. 14.2(b) is stretched by the same force F.
a)What is the maximum extension of the spring in the two cases ? (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ?
ANSWER-
For the one block system
Let mass m is extended by x When a force F is applied. For spring
F = k x Where, k is the spring constant.
∴x = F/K
This is maximum extension produced in one block system. And time period is given by
T = 2π√(m/k)
In case you are missed :- Previous Chapter Solution
14.14)
14.16) Answer the following questions:
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:
T= 2π√(m/k) .
A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2π√(L/9) Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
ANSWER-
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:
T= 2π√(m/k) .
But in case of simple pendulum k/m is constant and equal to w2. Hence the time period of a pendulum independent of the mass of the pendulum.
(b) we know that for simple pendulum the restoring force is given by,
F = mgsin (q)
And time period for small angle of oscillations time period is given by,
T = 2π√(L/9)
For small value of angle of oscillations sin(q) = q but if this angles of oscillation increases then sin (q) will be greater than (q) this will decrease the value of acceleration due to gravity which in turn increases the time period as time period is inversely proportional to the acceleration due to gravity.
(c) as wristwatch is working on the principle of spring and it is independent on the acceleration due to gravity hence The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall.
(d) When a simple pendulum mounted in a cabin falls freely under gravity its acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.
14.17) A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ?
Answer –
The cork is depressed slightly by distance x and then released. As a result, some extra water of a certain volume is displaced. Due to this an extra up-thrust acts upward and provides the restoring force to the cork.
∴ Restoring force( F) = Weight of the extra water displaced
14.19) One end of U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
ANSWER-
For column of mercury Restoring force is Weight of the mercury column of a certain height.
Force = mg
∴ F = mg
14.20) An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.14.27). Show that when the ball is pressed down a little and released , it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig. 14.27].
ANSWER-
when the ball is pressed down a little by distance x and released this would increase in the pressure inside the chamber decreases the volume of air.
This decrease in volume is given by,
V = ax
Bulk modulus of the air is given by
14.21)
14.22) Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
ANSWER-
The equation of displacement and velocity of a particle Performing SHM at time t is given by
From equation 1 and 2 we can conclude that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
14.23)
14.24)
14.25) A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation x = a cos (ωt+θ) and note that the initial velocity is negative.]
ANSWER-
GIVEN EQUATION
x = a cos (ωt+θ)
This is amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0.
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