The document Mechanical Oscillations (Part -1) - Oscillations and Waves, Irodov JEE Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.

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**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 v _{x}, and acceleration projection w_{x} as functions of time t; (b) velocity projection v_{x} and acceleration projection w_{x} as functions of the coordinate x. **

**Ans. 1. (a)** Given, x =

So

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

**(b)** From Eqn (1)

(2)

But from the law x = a cos (ωt - π/4) , so, x^{ 2} = a^{2} cos^{2} (ωt - π/4)

or, cos (ωt - π/4) - X^{2}/ = 2 or sin^{2} (ωt - 3t/4) = (3)

Using (3) in (2), (4)

Again from Eqn (4), w_{x} = -aω^{2}cos (ωt - π/4) = -ω^{2}x

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

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

(1)

Now compairing this equation with the general equation of harmonic oscillations : X - A sin(ωot+a) Amplitude, A = a/2 and angular frequency, ω_{0} - 2ω.

Thus the period of one full oscillation,

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

Plot of v_{x} (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 x_{o} = 25.0 cm and its velocity is equal to v_{x0 } = 100 cm/s. Find the coordinate x and the velocity v_{x} 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, v_{x} = - a **ω **sin (ωt + α)

Let us assume that at t = 0 , x = h_{0} and

Thus from Eqns (1) and (2) for t = 0, h_{0} = a cos α, and = - a **ω**sin α

Putting all the given numerical values, we get :

x = - 29 cm and v_{x} = - 81 cm/s

**Q. 4.Find the angular frequency and the amplitude of harmonic oscillations of a particle if at distances x _{1 } and x_{2} from the equilibrium position its velocity equals v_{1} a nd v_{2} respectively. **

** Ans. From the Eqn**

Solving these Eqns simultaneously, we get

**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 t_{x} be the time to cover the distence a/2 then from (1)

Hence sought mean velocity

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

x = a sin **ωt** (2)

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

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

Hence the sought mean velocity

**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 (v _{s}); (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.**

**(b) **In accordance with the problem

**(c) **We have got, v_{x} = a **ω** cos **ωt**

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

**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

Now one can write

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 t_{0} = 0).**Case (1)** when n is an odd number : In this case, from the equation

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

Thus the sought distance covered for odd n is

**Case (2)**, when n is even, In this case from the equation x = a cos **ω**t, the distance covered ( s_{2} ) in the interval t_{0}, is given by

or

Hence the sought distance for n is even

In general

**Q. 8. At the moment t = 0 a particle starts moving along the x axis so that its velocity projection varies as v _{x } = 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 v

Comparing v

Now we can write

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.

Hence the sought distance

**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

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

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.

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

We get

Note that

so

**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 _{1 } =3 .0 cos (ωt -1- π/3), x_{2} =8 .0 sin (ωt + π/6); (b) x_{1} = 3.0 cos ωt, x_{2} = 5.0 cos (ωt+ π/4), x_{3} =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

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

Then

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

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