NCERT Class 11 Physics Chapter 13 Oscillations Solutions, NCERT Class 11 Physics Chapter 13 Oscillations Notes to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 11 Physics Chapter 13 Oscillations Question Answer and select needs one.
NCERT Class 11 Physics Chapter 13 Oscillations
Also, you can read the SCERT book online in these sections NCERT Class 11 Physics Chapter 13 Oscillations Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. These solutions are part of SCERT All Subject Solutions. Here we have given NCERT Class 11 Physics Chapter 13 Oscillations Solutions for All Subjects, You can practice these here.
Oscillations
Chapter: 13
| Part – II |
EXERCISE
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.
Ans: (b) and (c) only.
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.
Ans: Periodic but not simple harmonic motion.
(b) Motion of an oscillating mercury column in a U-tube.
Ans: Simple harmonic motion.
(c) Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.
Ans: Simple harmonic motion.
(d) General vibrations of a polyatomic molecule about its equilibrium position.
Ans: Periodic but not simple harmonic motion.
3. Fig. 13.18 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)?
(a)
Ans: Does not represent periodic motion, as the motion neither repeats nor comes to mean position.
(b)
Ans: Represents the periodic motion, with periodic equal to 2s.
(c)
Ans: Does not represent periodic motion, because it is not identically repeated.
(d)
Ans: Represents periodic motion with periodic equals to 2s.
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.
Ans: Simple Harmonic.
(b) sin3 ωt.
Ans: Periodic but Non Simple Harmonic.
(c) 3 cos (π/4 – 2ωt).
Ans: Simple Harmonic.
(d) cos ωt + cos 3ωt + cos 5ωt.
Ans: Periodic but not Simple Harmonic.
(e) exp (–ω 2 t 2 ).
Ans: Non-Periodic.
(f) 1 + ωt + ω 2 t2.
Ans: Non-Periodic.
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:
Ans: A and B represent the two extreme positions of a S.H.M. for velocity, the direction from A to B, is taken to be positive. For acceleration and the force, the direction is taken positive if directed along AP and negative if directed along BP.
(a) At the end A.
Ans: Velocity = 0, Acceleration = Positive, Force = Positive.
(b) At the end B.
Ans: Velocity = 0, Acceleration = Negative, Force = Negative.
(c) at the midpoint of AB going towards A.
Ans: Velocity = Negative, Acceleration = Positive, Force = Positive.
(d) At 2 cm away from B going towards A.
Ans: Velocity = Negative, Acceleration = Positive, Force = Positive.
(e) At 3 cm away from A going towards B, and.
Ans: Velocity = Positive, Acceleration = Negative, Force = Negative.
(f) At 4 cm away from B going towards A.
Ans: Velocity = Positive, Acceleration = Negative, Force = Negative.
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 = –200𝓍2
(c) a = –10𝓍
(d) a = 100𝓍3.
Ans: Only (c) a = −10xa is the relationship that involves simple harmonic motion.
7. The motion of a particle executing simple harmonic motion is described by the displacement function,
x(t) = A cos (ωt + ∅).
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 s–1. 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.
Ans: At = t = 0, 𝓍 = 1 cm and 𝓋 = 𝜋 cm s-1
Φ = ? ω = 𝜋s-1
Given:
𝓍 = A cos (ωt + Φ)
Or: 1 = A cos (𝜋 × 0 + Φ)
= A cos Φ (i)
Velocity, V = dt/d𝓍 = – A ω sin (ωt + Φ)
or = – A𝜋 sin (𝜋 × 0 + Φ)
= – A𝜋 sinΦ
or 1 = -A sinΦ.
Hence, α = 3𝜋/4 + 𝜋/2 = 5𝜋/4.
8. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Ans: Here,
T = 2𝜋 √m/k
Given:
T = 0.6s
Scale reads up to 50kg over a length of 20 cm.
Spring constant K
= Fmax = 50 kg × 9.8 m/s2
= 490 N.
K = Fmax / Extension
= 490 N / 0.20 m
= 2450N/m
Finding the Mass m
T2 = (2𝜋)2 m/k
m = T2k / (2𝜋)2
m = (0.6)2 × 2450 / (2𝜋)2
m = 0.36 × 2450 / 4𝜋2
m = 882/ 39.478
m = 22.4 kg.
W (weight of the body):
W = m × g = 22.4 kg × 9.8 m/s2
= 219.52 N.
9. A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.
Determine:
(i) The frequency of oscillations.
Ans: Given:
Spring constant k = 1200 N/m
Mass, m = 3 kg
Displacement, 𝓍 = 2.0 cm = 0.02 m.
ω = √k/m
= √1200N/m / 3kg
√400 rad2/s2
= 20 rad/s
f = ω/ 2 𝜋
= 20 rad/s / 2𝜋
= 3.18 Hz.
(ii) Maximum acceleration of the mass. and
Ans: αmax = ω2 × 𝓍
αmax = (20 rad/s2) × 0.02 m
= 8 m/s2
(iii) The maximum speed of the mass.
Ans: 𝓋max = ω × 𝓍
𝓋max = 20 rad/s × 0.02m
= 0.4 m/s.
10. In Exercise 13.9, let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is:
(a) At the mean position.
Ans: If the time t = 0 at 𝓍 = 0 the displacement can be expressed as the sine function of t.
i.e., 𝓍(t) = A sin ωt
But, = √k/m = √1200/3
= 20 rad s-1
𝓍(t) = 2 sin 20 t.
(b) At the maximum stretched position. and
Ans: Mass at the maximum stretched position (𝓍 = A) at t = 0
At the maximum stretched position 𝓍 (0) = A.
This means cos (∅) = 1, so ∅ = 0.
The equation for 𝓍 (t) becomes:
𝓍(t) = Acos (ωt)
(c) At the maximum compressed position.
Ans: Mass at the Maximum compressed position ( 𝓍 = -A) at, t = 0
At the maximum stretched position 𝓍 (0) = A.
This means cos (∅) = 1, so ∅ = 0.
The equation for 𝓍 (t) becomes:
𝓍(t) = Acos (ωt + 𝜋) = – A cos (ωt)
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
Ans: Here the maximum displacement = Amplitude (A) = 2 cm.
11. Figures 13.20 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.Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
(a)
Ans: Let A be any point on the circle of reference of the image, From A, draw A perpendicular on 𝔁- axis,
If ∠POA = θ, then
∠OAM = θ, ωt
Now, in the triangle OAM,
OM/OA = sin θ
or = 𝔁 = -3 sin
Which is an equation of SHM.
(b)
Ans: Let B be any point on the circle of reference of the image. From B, draw BN,perpendicular on 𝔁 – axis.
Then ∠BON = θ = ωt
Now in ΔONB,
Cos θ = ON/OB
Or , ON = OB cos θ
Hence, -𝔁 = 2 cos ωt
Which is an equation of SHM.
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).
Ans: Amplitude (A): 2 cm
Angular frequency (ω): 3 rad/s
Initial phase (Ø): 𝜋/3
As we know that the function is negative, the particle starts at a position on the circle corresponding to – A sin (Ø), which is – 2 sin (𝜋/3) at t = 0.
(b) x = cos (π/6 – t)
Ans: Amplitude (A): 1 cm
Angular frequency (ω): 1 rad/s
Initial phase (Ø): 𝜋/6
(c) x = 3 sin (2πt + π/4)
Ans: Amplitude (A): 3 cm
Angular frequency (ω): 2 rad/s
Initial phase (Ø): 𝜋/4
(d) x = 2 cos πt.
Ans: Amplitude (A): 2 cm
Angular frequency (ω): 𝜋 rad/s
Initial phase (Ø): 0
The particle starts at 𝔁(0) = 2, which corresponds to the positive x-axis on the reference circle.
13. Figure 13.21 (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 13.21 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 13.21 (b) is stretched by the same force F.
(a)
(b)
(a) What is the maximum extension of the spring in the two cases?
Ans: Case (i) Spring clamped at one end with mass 𝓂 at the other end:
According to Hooke’s Law the extension of the spring 𝔁1
𝔁1 = F/K
Case (ii) Both ends of the spring free with mass 𝓂 at each and:
The force of the spring = 2F
According to Hooke’s Law the extension of the spring 𝔁2
𝔁2 = 2F/K.
(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?
Ans: Period of oscillation:
Case (i) Spring clamped at one end with mass 𝓂 at the other end:
T1 = 2𝜋 √m/k
Case (ii) Both ends of the spring free with mass 𝓂 at each and:
In this scenario, each mass 𝓂 contributes to the effective mass of the system.
T2 = 2𝜋√(m/2) /k
= 2𝜋 √m/2k.
14. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
Ans: Given that:
Stroke = 1.0 m
ω = 200 rad/min
We know that:
𝓋max = A × ω
Now find A:
A = Stroke/2
= 1.0m/2
= 0.5m
𝓋max = A × ω
= 0.5 m × 200 rad/min = 100m/min
Speed in m/s:
= 𝓋max = (100 m/min) / (60 s/min)
= 1.67 m/s.
15. The acceleration due to gravity on the surface of the moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s–2)
Ans: Here:
T = 2𝜋 √L/g
Given:
TEarth = 3.5s
gEarth = 9.8 m/s2
gMoon = 1.7 m/s2
TMoon/ TEarth = √gEarth/ gMoon
= TMoon / 3.5 = √9.8 / 1.7
TMoon = 3.5 × √9.8 / 1.7
TMoon = 3.5 × √5.76
TMoon = 3.5 × 2.4
TMoon = 8.4s
16. 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?
Ans: It in this case, the bob of the pendulum is under the action of two acceleration,
(i) Acceleration due to gravity, “g” acting vertically downwards.
(ii) Centripetal acceleration,
ac = 𝓋2/R acting along the horizontal direction.
Effective acceleration,
17. A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρl. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period.
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Ans: Let ɭ be the height of the cork which is inside the liquid. Then, at equilibrium, weight of the cork = upthrusts of the portion of the cork inside the liquid:
If the cork is slightly depressed below through a length of y, the net force acting in the upward directions is:
18. One end of a U-tube containing mercury is connected to a suction pump and the other end to the 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.
Ans: The restoring force,
F = weight of the mercury column of the height 2 ξ
F = -(volume) × density x g
F = – (A × 2ξ) × pg
= – 2Agpξ
Where,
A = Area of cross-section of the U-tube p = density of mercury.
Let L = length of the whole mercury column. Therefore, mass of mercury, M = Volume x density = A.L.p.
Where L is the total length of the mercury column.
or, L = 2h where ‘h’ is the height of the mercury column in U-tube. It shows that the mercury column executes S.Н.М.
