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Chapter 14 Oscillations Solutions

Question - 11 : - The following figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e., clockwise or anticlockwise) are indicated on each figure.
Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P in each case.

Answer - 11 : -


Question - 12 : -
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anti-clockwise in every case:
(x is in cm and t is in s)
(a) x = – 2 sin (3t + π /3)
(b) x = cos (π /6 – t)
(c) x = 3 sin (2πt + π /4)
(d) x = 2 cos π t.

Answer - 12 : -


Question - 13 : -
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Figure – (b) is stretched by the same force F.
 
(a) What is the maximum extension of the spring in the two cases?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released free, what is the period of oscillation in each case?

Answer - 13 : -


Question - 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 rev/min, what its maximum speed?

Answer - 14 : -


Question - 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 - 15 : -


Question - 16 : -

Answer - 16 : -


(c) The wristwatch uses an electronic system or spring system to give the time, which does not change with acceleration due to gravity. Therefore, watch gives the correct time.
(d) During free fall of the cabin, the acceleration due to gravity is zero. Therefore, the frequency of oscillations is also zero i.e., the pendulum will not vibrate at all.

Question - 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 - 17 : - In this case, the bob of the pendulum is under the action of two accelerations

Question - 18 : -
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ1 . 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 - 18 : -


Question - 19 : - One end of a U-tube containing mercury is connected to a suction pump and the other end to the 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 - 19 : -

The suction pump creates the pressure difference, thus mercury rises in one limb of the U-tube. When it is removed, a net force acts on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M. which can be expressed as:
Consider the mercury contained in a vertical U-tube upto the level P and Q in its two limbs.
Let P = density of the mercury.
L = Total length of the mercury column in both the limbs.
A = internal cross-sectional area of U-tube. m = mass of mercury in U-tube = LAP.
Assume, the mercury be depressed in left limb to F by a small distance y, then it rises by the same amount in the right limb to position Q’.
.’. Difference in levels in the two limbs = P’ Q’ = 2y.
:. Volume of mercury contained in the column of length 2y = A X 2y
.•. m – A x 2y x ρ.
If W = weight of liquid contained in the column of length 2y.
Then W = mg = A x 2y x ρ x g
This weight produces the restoring force (F) which tends to bring back the mercury to its equilibrium position.

Question - 20 : - An air chamber of volume V has a neck area of cross-section into which a ball of mass m just fits and can move up and down without any friction (Fig.). Shaw 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.

Answer - 20 : -

Consider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, Fig. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure.
Increase the pressure on the ball by a little amount p, so that the ball is depressed to position D, where CD = y.
There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV = Ay

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