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Page No 362: - Chapter 14 Oscillations class 11 ncert solutions Physics - SaraNextGen [2024]


Question 14.21:

You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant and (b) the damping constant for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

Answer:

(a) Mass of the automobile, m = 3000 kg

Displacement in the suspension system, x = 15 cm = 0.15 m

There are 4 springs in parallel to the support of the mass of the automobile.

The equation for the restoring force for the system:

F = –4kx = mg

Where, is the spring constant of the suspension system

Time period, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8098/Chapter%2014_html_m7b422208.gif

And https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8098/Chapter%2014_html_6d332a92.gif  = 5000 = 5 × 104 N/m

Spring constant, k = 5 × 104 N/m

(b) Each wheel supports a mass, M = https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8098/Chapter%2014_html_m266ef883.gif = 750 kg

For damping factor b, the equation for displacement is written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8098/Chapter%2014_html_6ccf9134.gif

The amplitude of oscillation decreases by 50%.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8098/Chapter%2014_html_7ca82b03.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8098/Chapter%2014_html_248a3ea0.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8098/Chapter%2014_html_64868c92.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8098/Chapter%2014_html_m2200c1a2.gif

Where,

Time period, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8098/Chapter%2014_html_509a9394.gif  = 0.7691 s

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8098/Chapter%2014_html_5a9871f3.gif = 1351.58 kg/s

Therefore, the damping constant of the spring is 1351.58 kg/s.

Question 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 of a particle executing SHM at an instant is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_57d26486.gif

Where,

A = Amplitude of oscillation

ω = Angular frequency https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_35fcc60e.gif

The velocity of the particle is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_7ab732dc.gif https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_21ee3236.gif

The kinetic energy of the particle is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_4e0c31a8.gif  https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_641c4427.gif

The potential energy of the particle is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_c4be639.gif https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_e8d104a.gif

For time period T, the average kinetic energy over a single cycle is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m2ebfea81.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m13b137ad.gif

And, average potential energy over one cycle is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4712/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m1e620bd8.gif

It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.

Question 14.23:

A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation = –α θ, where is the restoring couple and θ the angle of twist).

Answer:

Mass of the circular disc, m = 10 kg

Radius of the disc, r = 15 cm = 0.15 m

The torsional oscillations of the disc has a time period, T = 1.5 s

The moment of inertia of the disc is:

I https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8099/Chapter%2014_html_m7fcb8cb3.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8099/Chapter%2014_html_m5a4d85ce.gif × (10) × (0.15)2

= 0.1125 kg m2

Time period, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8099/Chapter%2014_html_m6b003488.gif

α is the torsional constant.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8099/Chapter%2014_html_6b501d62.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8099/Chapter%2014_html_54dcc2b2.gif

= 1.972 Nm/rad

Hence, the torsional spring constant of the wire is 1.972 Nm rad–1.

Question 14.24:

A body describes simple harmonic motion with amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.

Answer:

Amplitude, A = 5 cm = 0.05 m

Time period, T = 0.2 s

(a) For displacement, x = 5 cm = 0.05 m

Acceleration is given by:

a = –https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_4f423385.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_m4d82aad0.gif

Velocity is given by:

v

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_64f62dbf.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_m4eda7aa.gif

= 0

When the displacement of the body is 5 cm, its acceleration is –5π2 m/s2 and velocity is 0.

(b) For displacement, x = 3 cm = 0.03 m

Acceleration is given by:

=

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_c8c311f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_m7dd51187.gif

Velocity is given by:

v

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_64f62dbf.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_mcc232e1.gif

= 0.4 π m/s

When the displacement of the body is 3 cm, its acceleration is –3π m/s2 and velocity is 0.4π m/s.

(c) For displacement, x = 0

Acceleration is given by:

a

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_m684d41bf.gif = 0

Velocity is given by:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_m121a0afb.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5368/chapter%2014_html_68c74860.gif

When the displacement of the body is 0, its acceleration is 0 and velocity is 0.5 π m/s.

Question 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 = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation cos (ωt+θ) and note that the initial velocity is negative.]

Answer:

The displacement equation for an oscillating mass is given by:

x = https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5369/chapter%2014_html_m4611a111.gif

Where,

A is the amplitude

x is the displacement

θ is the phase constant

Velocity,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5369/chapter%2014_html_m72d2b09f.gif

At t = 0, x = x0

x0 = Acosθ = x0 … (i)

And,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5369/chapter%2014_html_m72f67260.gif https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5369/chapter%2014_html_5bfbe8be.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5369/chapter%2014_html_m3e8edca9.gif  … (ii)

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5369/chapter%2014_html_415e499f.gif

Hence, the amplitude of the resulting oscillation ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5369/chapter%2014_html_m33f04a40.gif .

Also Read : INTRODUCTION-Chapter-15-Waves-class-11-ncert-solutions-Physics

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