**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.

Let the equation of oscillation be given by
Velocity would be given as
Kinetic energy at an instant is given by
Time Period is given by
The Average Kinetic Energy would be given as follows
The potential energy at an instant T is given by
The Average Potential Energy would be given by
We can see Kav = Uav

**Q. 14.25 **A mass attached to a spring is free to oscillate, with angular velocity , in a horizontal plane without friction or damping. It is pulled to distance and pushed towards the centre with a velocity at time Determine the amplitude of the resulting oscillations in terms of the parameters , and . [Hint : Start with the equation and note that the initial velocity is negative.]

At the maximum extension of spring, the entire energy of the system would be stored as the potential energy of the spring.
Let the amplitude be A
The angular frequency of a spring-mass system is always equal to
Therefore

**Q. 14.24 (c)** A body describes simple harmonic motion with an amplitude of and a period of Find the acceleration and velocity of the body when the displacement is

(c)

A = 5 cm = 0.05 m
T = 0.2 s
At displacement x acceleration is
At displacement x velocity is
(a)At displacement 0 cm

**Q. 14.24 (b) **A body describes simple harmonic motion with an amplitude of and a period of Find the acceleration and velocity of the body when the displacement is

(b)

A = 5 cm = 0.05 m
T = 0.2 s
At displacement x acceleration is
At displacement x velocity is
(a)At displacement 3 cm

**Q. 14.24** **(a)** A body describes simple harmonic motion with an amplitude of and a period of Find the acceleration and velocity of the body when the displacement is

(a)

A = 5 cm = 0.05 m
T = 0.2 s
At displacement x acceleration is
At displacement x velocity is
(a)At displacement 5 cm

**Q .14.23 **A circular disc of mass 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 The radius of the disc is . Determine the torsional spring constant of the wire. (Torsional spring constant is defined by the relation, where J is the restoring couple and θ the angle of twist).

Moment of Inertia of the disc about the axis passing through its centre and perpendicular to it is
The period of Torsional oscillations would be

**Q. 14.21 (b)** You are riding in an automobile of mass . Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by during one complete oscillation. Estimate the values of

(b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports.

The amplitude of oscillation decreases by 50 % in one oscillation i.e. in one time period.
For damping factor b we have
x=x0/2
t=0.77s
m=750 kg

**Q. 14.21 (a) **You are riding in an automobile of mass . Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by during one complete oscillation. Estimate the values of

(a) the spring constant

Mass of automobile (m) = 3000 kg
There are a total of four springs.
Compression in each spring, x = 15 cm = 0.15 m
Let the spring constant of each spring be k

**Q. 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].

Let the initial volume and pressure of the chamber be V and P.
Let the ball be pressed by a distance x.
This will change the volume by an amount ax.
Let the change in pressure be
Let the Bulk's modulus of air be K.
This pressure variation would try to restore the position of the ball.
Since force is restoring in nature displacement and acceleration due to the force would be in different...

**Q. 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.

Let the height of each mercury column be h.
The total length of mercury in both the columns = 2h.
Let the cross-sectional area of the mercury column be A.
Let the density of mercury be
When either of the mercury columns dips by a distance x, the total difference between the two columns becomes 2x.
Weight of this difference is
This weight drives the rest of the entire column to the original...

**Q. 14.18 **A 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 is the density of cork. (Ignore damping due to viscosity of the liquid).

Let the cork be displaced by a small distance x in downwards direction from its equilibrium position where it is floating.
The extra volume of fluid displaced by the cork is Ax
Taking the downwards direction as positive we have
Comparing with a=-kx we have

**Q .14.17 **A simple pendulum of length l 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 . If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

Acceleration due to gravity = g (in downwards direction)
Centripetal acceleration due to the circular movement of the car = ac
(in the horizontal direction)
Effective acceleration is
The time period is T'

**Q. 14.16 (d) **Answer the following questions :

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

While free falling the effective value of g inside the cabin will be zero and therefore the frequency of oscillation of a simple pendulum would be zero i.e. it would not vibrate at all because of the absence of a restoring force.

**Q. 14.16 (c) **Answer the following questions :

(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?

The watch must be using an electrical circuit or a spring system to tell the time and therefore free falling would not affect the time his watch predicts.

**Q. 14.16 (b) **Answer the following questions :

(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.

In reaching the result we have assumed sin(x/l)=x/l. This assumption is only true for very small values of x . Therefore it is obvious that once x takes larger values we will have deviations from the above-mentioned value.

**Q. 14.16 (a) ** 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?

In case of spring, the spring constant is independent of the mass attached whereas in case of a pendulum k is proportional to m making k/m constant and thus the time period comes out to be independent of the mass of the body attached.

**Q. 14.15 **The acceleration due to gravity on the surface of moon is What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is ? (g on the surface of earth is 9.8 m s–2)

The time period of a simple pendulum of length l executing S.H.M is given by
ge = 9.8 m s-2
gm = 1.7 m s-2
The time period of the pendulum on the surface of Earth is Te = 3.5 s
The time period of the pendulum on the surface of the moon is Tm

**Q. 14.14** The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of If the piston moves with simple harmonic motion with an angular frequency of what is its maximum speed ?

Amplitude of SHM = 0.5 m
angular frequency is
If the equation of SHM is given by
The velocity would be given by
The maximum speed is therefore

**Q. 14.13 (b)** Figure 14.26 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. force applied at the free end stretches the spring. Figure 14.26 (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. 14.26 (b) is stretched by the same force .

(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?

(b).(a) In Fig, (a) we have
F=-kx
ma=-kx
(b) In fig (b) the two equal masses will be executing SHM about their centre of mass. The time period of the system would be equal to a single object of same mass m attached to a spring of half the length of the given spring (or undergoing half the extension of the given spring while applied with the same force)
Spring constant of such a spring would...

**Q. 14.13 (a) **Figure 14.30

(a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. force applied at the free end stretches the spring. Figure 14.30 (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. 14.30

(b) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases?

(a) Let us assume the maximum extension produced in the spring is x.
At maximum extension
(b) Let us assume the maximum extension produced in the spring is x. That is x/2 due to force towards left and x/2 due to force towards right

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