Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

JEE : Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

The document Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q.66. A coil of inductance L connects the upper ends of two vertical copper bars separated by a distance l. A horizontal conducting connector of mass m starts falling with zero initial velocity along the bars without losing contact with them. The whole system is located in a uniform magnetic field with induction B perpendicular to the plane of the bars. Find the law of motion x (t) of the connector. 

Ans. As the connector moves, an emf is set up in the circuit and a current flows, since the emf is
Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

provided x is measured from the initial position.

We then have

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

for by Lenz’s law the induced current will oppose downward sliding. Finally

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

on putting

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.67. A point performs damped oscillations according to the law Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev Find: (a) the oscillation amplitude and the velocity of the point at the moment t = 0; (b) the moments of time at which the point reaches the extreme positions. 

Ans. We are given Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

(a) The velocity of the point at t = 0 is obtained from

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

The term "oscillation amplitude at the moment t = 0" is meaningless. Probably the implication is the amplitude for Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev sin ωt and amplitude is ao.

(b) Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

when the displacement is an extremum. Then

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.68. A body performs torsional oscillations according to the law Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev Find: 

(a) the angular velocity Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev and the angular accelerationIrodov Solutions: Mechanical Oscillations- 6 Notes | EduRevof the body at the moment t = 0; (b) the moments of time at which the angular velocity becomes maximum. 

Ans. 

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.69. A point performs damped oscillations with frequency ω and damping coefficient   according to the law (4.1b). Find the initial amplitude a, and the initial phase a if at the moment t = 0 the displacement of the point and its velocity projection are equal to

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Ans. 

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 Since a0 is + ve, we must choose the upper sign if Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev and the lower sign if

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

(b) 

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.70. A point performs damped oscillations with frequency ω = = 25 s-1. Find the damping coefficient l if at the initial moment the velocity of the point is equal to zero and its displacement from the equilibrium position is η = 1.020 times less than the amplitude at that moment. 

Ans.

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.71. A point performs damped oscillations with frequency ω and damping coefficient β. Find the velocity amplitude of the point as a function of time t if at the moment t = 0
 (a) its displacement amplitude is equal to a0;
 (b) the displacement of the point x (0) = 0 and its velocity pro- jection 
Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Ans. 

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Velocity amplitude as a function of time is defined in the following manner. Put Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

then

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

forIrodov Solutions: Mechanical Oscillations- 6 Notes | EduRev This means that the displacement amplitude around the time Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev and we can say that the displacement amplitude at time Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev Similarly for the velocity amplitude.
Clearly

(a) Velocity amplitude at time  Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Since  Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

where y is anotner constant

(b)  Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

where a0 is real and positive.

Also

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Thus  Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev and we take - ( + ) sign if x0 is negative (positive). Finally the velocity amplitude is obtained as

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.72. There are two damped oscillations with the following periods T and damping coefficients β: T1 =  0.10 ms, β1  = 100s-1  and T2  = 10ms, β2  = 10s-1. Which of them decays faster?

Ans. The first oscillation decays faster in time. But if one takes the natural time scale, the period T for each oscillation, the second oscillation attenuates faster during that period. 

 

Q.73. A mathematical pendulum oscillates in a medium for which the logarithmic damping decrement is equal to 20  = 1.50. What will be the logarithmic damping decrement if the resistance of the medium increases n = 2.00 times? How many times has the resistance of the medium to be increased for the oscillations to become impossible? 

Ans. By definition of the logarithemic decrement Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRevwe get for the original decrement

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.74. A deadweight suspended from a weightless spring extends it by Δx = 9.8 cm. What will be the oscillation period of the dead Weight when it is pushed slightly in the vertical direction? The logarithmic damping decrement is equal to λ = 3.1.  

Ans.  The Eqn of the dead weight is 

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.75. Find the quality factor of the oscillator whose displacement amplitude decreases η = 2.0 times every n = 110 oscillations. 

Ans. The displacement amplitude decrease η times every n oscillations. Thus

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.76. A particle was displaced from the equilibrium position by a distance l = 1.0 cm and then left alone. What is the distance that the particle covers in the process of oscillations till the complete stop, if the logarithmic damping decrement is equal to λ = 0.020? 

Ans.

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

To get the maximum displacement in thfe second lap we note that

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

whereIrodov Solutions: Mechanical Oscillations- 6 Notes | EduRev, is the logarithemic decrement Substitution gives 2 metres.

 

Q.77. Find the quality factor of a mathematical pendulum l = = 50 cm long if during the time interval δ = 5.2 min its total mechanical energy decreases η = 4.0.10 times. 

Ans. For an undamped oscillator the mechanical energy Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRevis conserved. For a damped oscillator.

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

If Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev then the average of the last two terms over many oscillations about the time t will vanish and

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

and this is the relevant mechanical energy.

In time δ this decreases by a factor Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRevso

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.78. A uniform disc of radius R = 13 cm can rotate about a horizontal axis perpendicular to its plane and passing through the edge of the disc. Find the period of small oscillations of that disc if the logarithmic damping decrement is equal to λ = 1.00. 

Ans. The restoring couple is

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

The moment of inertia is

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Thus for undamped oscillations

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.79. A thin uniform disc of mass m and radius R suspended by an elastic thread in the horizontal plane performs torsional oscillations in a liquid. The moment of elastic forces emerging in the thread is equal to N = αφ, where a is a constant and IT is the angle of rotation from the equilibrium position. The resistance force acting on a unit area of the disc is equal to F1 = ηv, where is a constant and η is the velocity of the given element of the disc relative to the liquid. Find the frequency of small oscillation. 

Ans. Let us calculate the moment G1 of all the resistive forces on the disc. When the disc rotates an element (rdrdθ) with coordinates (r,θ) has a velocity rIrodov Solutions: Mechanical Oscillations- 6 Notes | EduRev, where φ is the instantaneous angle of rotation from the equilibrium position and r is measured from the centre. Then

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Also moment of inertia =Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

and angular frequency Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Note:- normally by frequency we mean Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

 

Q.80. A disc A of radius R suspended by an elastic thread between two stationary planes (Fig. 4.24) performs torsional oscillations about its axis 00'. The moment of inertia of the disc relative to that axis is equal to I, the clearance between the disc and each of the planes is equal to h, with h << R. Find the viscosity of the gas surrounding the disc A if the oscillation period of the disc equals T and the logarithmic damping decrement, λ. 

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Ans. From the law of viscosity, force per unit area = Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

so when the disc executes torsional oscillations the resistive couple on it is

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

(factor 2 for the two sides of the disc; see the figure  in the book) where φ is torsion. The equation of motion is

Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

Now the logarithmic decrement Irodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

ThusIrodov Solutions: Mechanical Oscillations- 6 Notes | EduRev

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