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Page No 359: - Chapter 14 Oscillations class 11 ncert solutions Physics - SaraNextGen [2024]


Question 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 ω+ cos 3ωt + cos 5ωt

(e) exp (–ω2t2)

(f) 1 + ωt ω2t2

Answer:

(a) SHM

The given function is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_2938ee21.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m76ee01f8.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_646e67b1.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_ma66a894.gif

This function represents SHM as it can be written in the form:https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m70d916e8.gif

Its period is: https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_7ccdb634.gif

(b) Periodic, but not SHM

The given function is:

sin3ωt=143sinωt-sin3ωt

The terms sin ωt and sin ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.

(c) SHM

The given function is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m404087ce.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_471b9cdd.gif

This function represents simple harmonic motion because it can be written in the form:https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_23786791.gif

Its period is: https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_1b4eae3c.gif

(d) Periodic, but not SHM

The given function is https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_mde3c013.gif . Each individual cosine function represents SHM. However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.

(e) Non-periodic motion

The given function https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4694/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m7583035a.gif  is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.

(f) The given function 1 + ωt + ω2t2 is non-periodic.

Question 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:

Answer:

(a) Zero, Positive, Positive

(b) Zero, Negative, Negative

(c) Negative, Zero, Zero

(d) Negative, Negative, Negative

(e) Positive, Positive, Positive

(f) Negative, Negative, Negative

Explanation:

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.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4695/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_636e2f34.jpg

A particle is in linear simple harmonic motion between the end points

(a) At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this point.

Its acceleration is positive as it is directed along AO.

Force is also positive in this case as the particle is directed rightward.

(b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point.

Its acceleration is negative as it is directed along O.

Force is also negative in this case as the particle is directed leftward.

(c)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4695/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m2ff65307.jpg

The particle is executing a simple harmonic motion. O is the mean position of the particle. Its velocity at the mean position O is the maximum. The value for velocity is negative as the particle is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position.

(d)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4695/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_3b435578.jpg

The particle is moving toward point O from the end B. This direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, the particle’s velocity and acceleration, and the force on it are all negative.

(e)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4695/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_3c952cab.jpg

The particle is moving toward point O from the end A. This direction of motion is from A to B, which is the conventional positive direction. Hence, the values for velocity, acceleration, and force are all positive.

(f)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4695/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m22fd13ee.jpg

This case is similar to the one given in (d).

Question 14.6:

Which of the following relationships between the acceleration and the displacement of a particle involve simple harmonic motion?

(a) = 0.7x

(b) = –200x2

(c) = –10x

(d) = 100x3

Answer:

(c) A motion represents simple harmonic motion if it is governed by the force law:

F = –kx

ma = –k

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4696/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m5f56f94.gif

Where,

F is the force

m is the mass (a constant for a body)

x is the displacement

a is the acceleration

k is a constant

Among the given equations, only equation a = –10 is written in the above form with https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4696/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_3058c02e.gif  Hence, this relation represents SHM.

Question 14.7:

The motion of a particle executing simple harmonic motion is described by the displacement function,

x (t) = cos (ω+ φ).

If the initial (= 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.

Answer:

Initially, at t = 0:

Displacement, x = 1 cm

Initial velocity, v = ω cm/sec.

Angular frequency, ω = π rad/s–1

It is given that:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4697/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m55127882.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4697/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_1f79204b.gif

Squaring and adding equations (i) and (ii), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4697/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_2da391a8.gif

Dividing equation (ii) by equation (i), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4697/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m44fab2e8.gif

SHM is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4697/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m6624ba82.gif

Putting the given values in this equation, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4697/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_2a5fc6e4.gif

Velocity, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4697/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m85480b4.gif

Substituting the given values, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4697/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_67ff4d53.gif

Squaring and adding equations (iii) and (iv), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4697/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_35e716a4.gif

Dividing equation (iii) by equation (iv), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4697/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_56fd8abe.gif

Question 14.8:

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?

Answer:

Maximum mass that the scale can read, M = 50 kg

Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m

Time period, T = 0.6 s

Maximum force exerted on the spring, F = Mg

Where,

g = acceleration due to gravity = 9.8 m/s2

= 50 × 9.8 = 490

∴Spring constant,https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4698/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_db85862.gif  https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4698/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_425ce53b.gif

Mass m, is suspended from the balance.

Time period, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4698/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m6282bf2c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4698/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_565a3f63.gif

∴Weight of the body = mg = 22.36 × 9.8 = 219.167 N

Hence, the weight of the body is about 219 N.

Question 14.9:

A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 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.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4699/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m7f9deb63.jpg

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Answer:

Spring constant, k = 1200 N m–1

Mass, = 3 kg

Displacement, A = 2.0 cm = 0.02 cm

(i) Frequency of oscillation v, is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4699/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_97a7778.gif

Where, T is the time period

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4699/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_3e2b66d5.gif

Hence, the frequency of oscillations is 3.18 cycles per second.

(ii) Maximum acceleration (a) is given by the relation:

a = ω2 A

Where,

ω = Angular frequency = https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4699/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m47782030.gif

A = Maximum displacement

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4699/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_6a3ba554.gif

Hence, the maximum acceleration of the mass is 8.0 m/s2.

(iii) Maximum velocity, vmax = Aω

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4699/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m46aa47e7.gif

Hence, the maximum velocity of the mass is 0.4 m/s.

Question 14.10:

In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give as a function of time t for the oscillating mass if at the moment we start the stopwatch (= 0), the mass is

(a) at the mean position,

(b) at the maximum stretched position, and

(c) at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer:

(a) x = 2sin 20t

(b) x = 2cos 20t

(c) x = –2cos 20t

The functions have the same frequency and amplitude, but different initial phases.

Distance travelled by the mass sideways, A = 2.0 cm

Force constant of the spring, k = 1200 N m–1

Mass, m = 3 kg

Angular frequency of oscillation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8095/Chapter%2014_html_m3126f34b.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8095/Chapter%2014_html_m443ec16a.gif  = 20 rad s–1

(a) When the mass is at the mean position, initial phase is 0.

Displacement, x = Asin ωt

2sin 20t

(b) At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8095/Chapter%2014_html_m4c4df2e.gif .

Displacement,https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8095/Chapter%2014_html_b30eb56.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8095/Chapter%2014_html_55ef364c.gif

= 2cos 20t

(c) At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8095/Chapter%2014_html_1549eb8f.gif .

Displacement,https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8095/Chapter%2014_html_5d88fe5e.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8095/Chapter%2014_html_m680c200.gif  = –2cos 20t

The functions have the same frequencyhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8095/Chapter%2014_html_43c92651.gif  and amplitude (2 cm), but different initial phaseshttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8095/Chapter%2014_html_m32cfc76c.gif .

Also Read : Page-No-360:-Chapter-14-Oscillations-class-11-ncert-solutions-Physics

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