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Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE PDF Download

Q. 1. A point oscillates along the x axis according to the law x = a cos (ωt — n/4). Draw the approximate plots
 (a) of displacement x, velocity projection vx, and acceleration projection wx  as functions of time t;
 (b) velocity projection vx  and acceleration projection wx  as functions of the coordinate x. 

Ans. 1. (a) Given, x =  Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

So Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

On-the basis of obtained expressions plots x(t) , vx(t) and wx(t) can be drawn as shown in the answersheet, (of the problem book ). 

(b) From Eqn (1)

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE   (2)

But from the law x = a cos (ωt - π/4) , so, x 2 = a2 cos2 (ωt - π/4)

or, cos (ωt - π/4) - X2/ = 2 or sin2 (ωt - 3t/4) = Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE   (3)

Using (3) in (2), Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE    (4)

Again from Eqn (4), wx = -aω2cos (ωt - π/4) = -ω2x

 

Q. 2. A point moves along the x axis according to the law x = a sin2(ωt — π/4). Find: (a) the amplitude and period of oscillations; draw the plot x (t); (b) the velocity projection vx  as a function of the coordinate x; draw the plot vx  (x).

Ans. 2. (a) From the motion law of the particle

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE 
                                                                               (1)

Now compairing this equation with the general equation of harmonic oscillations : X - A sin(ωot+a) Amplitude, A = a/2 and angular frequency, ω0 - 2ω.

Thus the period of one full oscillation, Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

(b) Differentiating Eqn (1) w.r.t. time

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Plot of vx (x) is as shown in the answersheet.

 

Q. 3.A particle performs harmonic oscillations along the x axis about the equilibrium position x = 0. The oscillation frequency is  ω = 4.00 s-1 . At a certain moment of time the particle has a coor- dinate xo  = 25.0 cm and its velocity is equal to vx0  = 100 cm/s. Find the coordinate x and the velocity vx of the particle t = 2.40s after that moment.

 Ans. 3 Let the general equation of S.H.M. be
h = a cos (ωt + α)
So, vx = - a ω sin (ωt + α)
Let us assume that at t = 0 , x = h0 and Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus from Eqns (1) and (2) for t = 0, h0 = a cos  α, and  Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE= - a ωsin α

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Putting all the given numerical values, we get :

x = - 29 cm and vx = - 81 cm/s 

 

Q. 4.Find the angular frequency and the amplitude of harmonic oscillations of a particle if at distances x and x2  from the equilibrium position its velocity equals v1 a nd v2  respectively. 

 Ans.  From the Eqn

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Solving these Eqns simultaneously, we get

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q. 5. A point performs harmonic oscillations along a straight line with a period T = 0.60 s and an amplitude a = 10.0 cm. Find the mean velocity of the point averaged over the time interval during which it travels a distance a/2, starting from
 (a) the extreme position;
 (b) the equilibrium position. 

 Ans.  (a) When a particle starts from an extreme position, it is useful to write the motion law as x = a cos ωt             (1)

(However x is the displacement from the equlibrium position)

It tx be the time to cover the distence a/2 then from (1)

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence sought mean velocity

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

(b) In this case, it is easier to write the motion law in the form :

x = a sin ωt                       (2)

If t2 be the time to cover the distance a/2, then from Eqn (2)

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Differentiating Eqn (2) w.r.t time, we get

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence the sought mean velocity

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q. 6. At the moment t = 0 a point starts oscillating along the x axis according to the law x = a sin ωt. Find:
 (a) the mean value of its velocity vector projection (vs);
 (b) the modulus of the mean velocity vector |(v)| ;
 (c) the mean value of the velocity modulus (v) averaged over 3/8 of the period after the start.  

 Ans. (a) As x = a sin ωt                  so,   vx = aω cos ωt

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

(b) In accordance with the problem

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

(c) We have got, vx = a ω cos ωt

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Using ω = 2 ω/T , and on evaluating the integral we get

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q. 7. A particle moves along the x axis according to the law x = a cos ωt. Find the distance that the particle covers during the time interval from t = 0 to t.

 Ans. From the motion law, x = a cos ωt„ it is obvious that the time taken to cover the distance equal to the amplitude (a), starting from extreme position equals T/4.
Now one can write

 Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

As the particle moves according to the law, x .= a cos ωt, so at n = 1,3,5 ... or for odd n values it passes through the mean positon and for even numbers of n it comes to an extreme position (if t0 = 0).
Case (1) when n is an odd number : In this case, from the equation

x = x a sin ωt, if the t is counted from nT/4 and the distance covered in the time interval to be comes  

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus the sought distance covered for odd n is

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Case (2), when n is even, In this case from the equation x = a cos ωt, the distance covered ( s2 ) in the interval t0, is given by

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

or  Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence the sought distance for n is even

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

In general

 Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q. 8. At the moment t = 0 a particle starts moving along the x axis so that its velocity projection varies as v = 35 cos πt cm/s, where t is expressed in seconds. Find the distance that this particle covers during t = 2.80 s after the start. 

Ans. Obviously the motion law is of the from, x = a shoot, and vx = ω a cos ωt.
Comparing vx - ω a cos ωt with vx - 35 cos πt , we get

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Now we can write

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

As n = 5 is odd, like (4 = 7), we have to basically find the distance covered by the particle starting from the extreme position in the time interval 0 = 3 s.
Thus from the Eqn.

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence the sought distance

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q. 9. A particle performs harmonic oscillations along the x axis according to the law x = a cos ωt. Assuming the probability P of the particle to fall within an interval from —a to +a to be equal to unity, find how the probability density dP/dx depends on x. Here dP denotes the probability of the particle falling within an interval from x to x dx. Plot dP/dx as a function of x. 

Ans.  As the motion is periodic the particle repeatedly passes through any given region in the range - a ≤ x ≤ a.

The probability that it lies in the range (x, x + dx) is defined as the fraction

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE where Δt is the time that the particle lies in the range (x, x + dx) out of the total time t. Because of periodicity this is

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

where the factor 2 is needed to take account of the fact that the particle is in the range ( x , x + d x ) during both up and down phases of its motion. Now in a harmonic oscillator.

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus since ωT = 2 π ( T is the time period)

We getIrodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Note that Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

so Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Q. 10. Using graphical means, find an amplitude a of oscillations resulting from the superposition of the following oscillations of the same direction:
 (a) x =3 .0 cos (ωt -1- π/3), x2  =8 .0 sin (ωt + π/6);
 (b) x1 =  3.0 cos ωt, x2  = 5.0 cos (ωt+ π/4), x3  =6 .0 sin ωt. 

Ans.  (a) We take a graph paper and choose an axis (X - axis) and an origin. Draw a vector of magnitude 3 inclined at an angle π/3 with the X -axis. Draw another vector of magnitude 8 inclined at an angle - π/3( Since sin (ωt+ π/ 6 ) » cos (ωt- π/ 3 )) with the X - axis. The magnitude of the resultant of both these vectors (drawn from the origin) obtained using parallelogram law is the resultant, amplitude.

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

 

Thus            R = R7 units

(b) One can follow the same graphical method here but the result can be obtained more quickly by breaking into sines and cosines and adding :
Resultant

Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Then
Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

So,   A = 6-985  = 7 units
Note- In using graphical method convert all oscillations to either sines or cosines but do not use both.

The document Irodov Solutions: Mechanical Oscillations- 1 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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FAQs on Irodov Solutions: Mechanical Oscillations- 1 - I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

1. What is mechanical oscillation?
Ans. Mechanical oscillation refers to the repetitive back-and-forth movement of a mechanical system about its equilibrium position. It is characterized by a periodic variation in displacement, velocity, and acceleration.
2. What are the different types of mechanical oscillations?
Ans. There are various types of mechanical oscillations, including simple harmonic motion, damped oscillations, forced oscillations, and coupled oscillations. Each type exhibits different characteristics and behaviors.
3. How is simple harmonic motion related to mechanical oscillations?
Ans. Simple harmonic motion is a type of mechanical oscillation where the restoring force acting on a system is directly proportional to the displacement from the equilibrium position and is always directed towards the equilibrium point. Many physical systems exhibit simple harmonic motion, such as a mass-spring system or a pendulum.
4. What factors affect the frequency of mechanical oscillations?
Ans. The frequency of mechanical oscillations depends on factors such as the stiffness of the system, the mass involved, and any external forces acting on the system. Increasing the stiffness or decreasing the mass will result in a higher frequency, while external forces can alter the natural frequency of the system.
5. How can mechanical oscillations be used in practical applications?
Ans. Mechanical oscillations find practical applications in various fields. For example, they are utilized in the design of musical instruments, such as guitars and pianos, to produce specific frequencies and notes. They are also used in timekeeping devices like pendulum clocks and in engineering applications such as vibration analysis and control in machinery.
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