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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
Onthe 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 ωsin 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. (a) As x = a sin ωt so, v_{x} = aω cos ωt
(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 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
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 ωt, 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 ωt, 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_{x} = ω a cos ωt.
Comparing v_{x}  ω a cos ωt with v_{x}  35 cos πt , we get
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 ωT = 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 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.
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 = 6985 = 7 units
Note In using graphical method convert all oscillations to either sines or cosines but do not use both.
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