Page No 361: - Chapter 14 Oscillations class 11 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

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,

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

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? (g on the surface of earth is 9.8 ms^–2)

Answer:

Acceleration due to gravity on the surface of moon,  = 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

Where,

l is the length of the pendulum

The length of the pendulum remains constant.

On moon’s surface, time period, 

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 k and mass m of the particle:

. 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 . 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, 

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

k  m

 = 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,

F = Restoring force

m = Mass of the bob

g = Acceleration due to gravity

θ = Angle of displacement

For small θ, sinθ 

For large θ, sinθ is greater than θ.

This decreases the effective value of g.

Hence, the time period increases as:

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 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 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 

Where,

v is the uniform speed of the car

R is the radius of the track

Effective acceleration (a_eff) is given as:

Time period, 

Where, l is the length of the pendulum

∴Time period, T 

Question 14.18:

Cylindrical piece of cork of density of base area A and height h floats in a liquid of density . 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).

Answer:

Base area of the cork = A

Height of the cork = h

Density of the liquid =

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 x   g … (i)

According to the force law:

F = kx

Where, k is a constant

The time period of the oscillations of the cork:

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:

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) = –2Aρgh = –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 

Time period, 

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

∴

Hence, the mercury column executes simple harmonic motion with time period .

Question 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.14.33). 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].

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 x 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 

Bulk Modulus of air,

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

The restoring force acting on the ball,

F = p × a

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:

Time period,