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Page No 361: - Chapter 14 Oscillations class 11 ncert solutions Physics - SaraNextGen [2024]


Question 14.14:

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?

Answer:

Angular frequency of the piston, ω = 200 rad/ min.

Stroke = 1.0 m

Amplitude,https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4704/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_514b26a0.gif

The maximum speed (vmax) of the piston is give by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4704/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_65a85ba1.gif

Question 14.15:

The acceleration due to gravity on the surface of moon is 1.7 ms–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? (on the surface of earth is 9.8 ms–2)

Answer:

Acceleration due to gravity on the surface of moon, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4705/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m461d7dd3.gif = 1.7 m s–2

Acceleration due to gravity on the surface of earth, g = 9.8 m s–2

Time period of a simple pendulum on earth, T = 3.5 s

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4705/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m16a7ae30.gif

Where,

l is the length of the pendulum

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4705/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_3ccdf4dd.gif

The length of the pendulum remains constant.

On moon’s surface, time period, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4705/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m34a0f79e.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4705/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m34756763.gif

Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.

Question 14.16:

Answer the following Questions:

(a) Time period of a particle in SHM depends on the force constant and mass of the particle:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5370/chapter%2014_html_m6282bf2c.gif . 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 is greater thanhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5370/chapter%2014_html_m69883a91.gif . 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 cabinthat is freely falling under gravity?

Answer:

(a) The time period of a simple pendulum, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5370/chapter%2014_html_m6282bf2c.gif

For a simple pendulum, k is expressed in terms of mass m, as:

k https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5370/chapter%2014_html_10490900.gif m

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5370/chapter%2014_html_m6157e532.gif  = Constant

Hence, the time period T, of a simple pendulum is independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:

F = –mg sinθ

Where,

= Restoring force

m = Mass of the bob

g = Acceleration due to gravity

θ = Angle of displacement

For small θ, sinθ https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5370/chapter%2014_html_3b821a1f.gif

For large θ, sinθ is greater than θ.

This decreases the effective value of g.

Hence, the time period increases as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/5370/chapter%2014_html_6ddff4d9.gif

Where, l is the length of the simple pendulum

(c) The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall. Since a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. Its working depends on spring action.

(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.

Question 14.17:

A simple pendulum of length and having a bob of mass is suspended in a car. The car is moving on a circular track of radius 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 bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.

Acceleration due to gravity = g

Centripetal acceleration https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4707/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m311237ff.gif

Where,

v is the uniform speed of the car

R is the radius of the track

Effective acceleration (aeff) is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4707/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m6f6414d6.gif

Time period, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4707/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m1c73f895.gif

Where, l is the length of the pendulum

∴Time period, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4707/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_120cfc22.gif

Question 14.18:

Cylindrical piece of cork of density of base area and height floats in a liquid of densityhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4708/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m5f01a1f6.gif . The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4708/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m41febef.gif

where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Answer:

Base area of the cork = A

Height of the cork = h

Density of the liquid =

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4708/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m5f01a1f6.gif

Density of the cork = ρ

In equilibrium:

Weight of the cork = Weight of the liquid displaced by the floating cork

Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.

Up-thrust = Restoring force, F = Weight of the extra water displaced

F = ­–(Volume × Density × g)

Volume = Area × Distance through which the cork is depressed

Volume = Ax

∴ F = – A https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4708/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m5f01a1f6.gif  g … (i)

According to the force law:

F = kx

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4708/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m7187d34b.gif

Where, k is a constant

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4708/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m37be4f02.gif

The time period of the oscillations of the cork:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4708/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m3fc1a9ec.gif

Where,

m = Mass of the cork

= Volume of the cork × Density

= Base area of the cork × Height of the cork × Density of the cork

Ahρ

Hence, the expression for the time period becomes:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4708/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_2d3a6634.gif

Question 14.19:

One end of a 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:

Area of cross-section of the U-tube = A

Density of the mercury column = ρ

Acceleration due to gravity = g

Restoring force, F = Weight of the mercury column of a certain height

F = –(Volume × Density × g)

F = –(A × 2h × ρ ×g) = –2gh = –k × Displacement in one of the arms (h)

Where,

2h is the height of the mercury column in the two arms

k is a constant, given by https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4709/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m811ce11.gif

Time period, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4709/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_m6282bf2c.gif https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4709/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_2897da8c.gif

Where,

m is the mass of the mercury column

Let l be the length of the total mercury in the U-tube.

Mass of mercury, m = Volume of mercury × Density of mercury

Alρ

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4709/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_286f006d.gif

Hence, the mercury column executes simple harmonic motion with time periodhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/4709/NS_3-10-08_Sravana_11_Physis_14_25_NRJ_html_258767c4.gif .

Question 14.20:

An air chamber of volume has a neck area of cross section into which a ball of mass just fits and can move up and down without any friction (Fig.14.33). 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. 14.33].

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8097/Chapter%2014_html_342f5d9c.jpg

Answer:

Volume of the air chamber = V

Area of cross-section of the neck = a

Mass of the ball = m

The pressure inside the chamber is equal to the atmospheric pressure.

Let the ball be depressed by units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber.

Decrease in the volume of the air chamber, ΔV = ax

Volumetric strain https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8097/Chapter%2014_html_78d4ce7a.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8097/Chapter%2014_html_195f3b77.gif

Bulk Modulus of air,https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8097/Chapter%2014_html_m6afec234.gif

In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8097/Chapter%2014_html_1a2f5ba1.gif

The restoring force acting on the ball,

F = p × a

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8097/Chapter%2014_html_217f650d.gif

In simple harmonic motion, the equation for restoring force is:

F = –kx … (ii)

Where, k is the spring constant

Comparing equations (i) and (ii), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8097/Chapter%2014_html_m7979b771.gif

Time period,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8097/Chapter%2014_html_m6282bf2c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/192/8097/Chapter%2014_html_5195506d.gif

Also Read : Page-No-362:-Chapter-14-Oscillations-class-11-ncert-solutions-Physics

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