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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Friday, December 21, 2012

### AP Physics B - Multiple Choice Practice Questions on Doppler effect

In the post dated 9th April 2009 the essential points to be noted about Doppler effect were given. You may click here to access that post. All posts related to Doppler effect can be accessed by clicking on the label ‘Doppler effect’ below this post.

Today we shall discuss a few multiple choice practice questions involving Doppler effect:

(1) The figure above shows a car sounding its horn and moving with uniform velocity along a straight road. At time t = 0 the car is at position A and at time t = t1 the car moves past a stationary listener L. At time t = t2 the car is at position B. Which one among the following graphs best represents the variation of the frequency (pitch) of the horn with time as heard by the listener?

When the car moves towards the listener the number of complete sound waves reaching the ear of the listener in one second is greater than the number reaching his ears when the car is at rest. The frequency of the horn as heard by the listener in this case is higher than the real requency of the horn. When the car moves away from the listener the number of complete sound waves reaching the ears of the listener in one second is less than the number of waves produced by the horn in one second.

[All the sound waves produced by the horn in one second.cannot reach the listener’s ears within one second since the car is moving away].

Therefore, the frequency of the horn as heard by the listener is less than the real frequency after the car moves past the listener.

The graph which best represents the variation of the frequency with time is (a).

(2) In the above question suppose the car is moving with uniform acceleration. Which one among the graphs will best represent the variation of the frequency (pitch) of the horn with time as heard by the listener?

In this case the pitch of the sound as heard by the listener will go on increasing when the car moves towards the listener since the relative velocity of the car with respect to the listener goes on increasing. After the car moves past the listener the pitch of the sound as heard by the listener will go on decreasing since the source of sound (horn) is moving away from the listener with continuously increasing velocity.

Therefore, the graph which best represents the variation of the frequency with time in this case is (e).

(3) A stationary police car at position A (Fig.) sounds its horn when the air is still and a listener L on the road hears the frequency of the note as n. A steady wind now blows continupusly from B to A with speed equal to 2 % of the speed of sound in air. What will be the frequency of the sound of  the siren as heard by the listener when the wind is blowing?

(a) 0.98 n

(b) 1.02 n

(c) n

(d) 0.99 n

(e) 1.01 n

The number of complete sound waves emitted per second by the siren will be n whether the wind is present or absent. Since the steady wind is blowing continuously a stationary listener will receive just n complete waves per second. The correct option is (c).

[If the sound source and listener are moving relative to each other, the listener will hear one shifted frequency (Doppler shift) in still air and a different shifted frequency if wind is blowing. However, if the source and the listener are at rest relative to each other, there is no Doppler shift when the wind is blowing. If you are confused, you may use the expression for the apparent frequency as heard by the listener:

Let the source S of sound (fig.) move with velocity vS, the listener move with velocity vL and the wind blow with velocity vw, all in the same direction as shown in the figure.

Wind→ w         S●→vS        L●→vL

The apparent frequency (n’) of sound is then given by

n’ = n(v + w vL)/ (v + w vS)

where n is the real frequency of the sound and v is the velocity of sound.

If vL and vS are zero we obtain n’ = n

When there is relative motion between the source and the listener, you will have to take the sign of the wind velocity w as positive, as given in the above expression, when the wind blows from source to the listener. If the wind blows from listener to source, the sign of w is to be taken as negative]

You will find more questions (with solution) in this section at

## Friday, November 16, 2012

### AP Physics C - Multiple Choice Practice Questions on One Dimensional Motion

“Maturity is often more absurd than youth and very frequently is most unjust to youth.”
Thomas A. Edison

Today’s post covers a few multiple choice practice questions related to one dimensional kinematics meant for AP Physics C aspirants.
(1) Water drops from a leaking overhead tank falls on the ground 5 m below, at regular intervals, the 11th  drop just beginning to fall when the first drop strikes the ground. What will be the height of the 9th drop when the first drop strikes the ground? (Acceleration due to gravity = 10 ms–2)
(a) 4.9 m
(b) 4.8 m
(c) 4.6 m
(d) 4.4 m
(e) 4.2 m

The time t taken by a drop to fall through 5 m is given by the relevant equation of motion,
5 = 0 + ½ gt2
Since g = 10 ms–2 the above equation gives t = 1 second
When the first drop strikes the ground after falling freely for one second, the 9th  drop has fallen freely for one-fifth of one second, as is evident from the adjoining figure (remembering that the drops fall at regular intervals of time). Therefore, the 9th drop has fallen through a distance s given by
s = 0 + ½ g(0.2)2 = (½)×10×0.04 = 0.2 m
Therefore the height of the 9th drop when the first drop strikes the ground is 5 m – 0.2 m = 4.8 m.
(2) An object projected vertically upwards with initial velocity u attains maximum height in 5 s. The ratio of the distance traveled by the object in the 1st second and the 6th second is (Acceleration due to gravity = 10 ms–2)
(a) 6 : 1
(b) 8 : 1
(c) 9 : 1
(d) 10: 1
(e) 11 : 1
The velocity of projection (u) of the object is given by the equation of uniformly accelerated linear motion,
0 = u gt
[We have used the equation, v = u + at with usual notations].
Substituting for g and t we have
0 = u – 10×5
Therefore u = 50 ms–1
The distance s1 traveled by the object in the 1st second is given by
s1 = (50×1) – (½ ×10×12) = 45 m
[We have used the equation, s = ut + ½ at2 with usual notations]
At the end of 5 seconds the object is at the highest point of its trajectory where its velocity is zero. Therefore, the distance s6 traveled by the object during the next one second (6th second) is given by
s6 = 0 + ½ ×10×12 = 5 m.
[We have used the equation of motion, s = ut + ½ gt2 with usual notations]
The ratio of the distance traveled by the object in the 1st second and the 6th second is
s1 : s6 = 45 : 5 = 9 : 1
(3) A ball projected vertically upwards with initial velocity u reaches maximum height h in t sec. What is the total time taken by the ball (from the instant of projection) to reach a height h/4 while returning?
(a) 1.75t
(b) 1.65t
(c) 1.5t
(d) 1.4t
(e) 1.3t
From the equation of motion, v2 = u2 + 2as with usual notations, we have for the upward journey
0 = u2 2gh
Therefore h = u2/2g
From the equation of motion, v = u + at with usual notations, we have for the upward journey
0 = u gt so that u = gt
Substituting for u in the expression for h we have
h = gt2/2 …………. (i)
We now use the equation of motion, s = ut + ½ gt2 for the free fall of the ball from the maximum height h to the height h/4.
Since distance of fall is 3h/4) we have
3h/4 = 0 + ½ gt12 where t1 is the time of fall from the maximum height h to the height h/4.
Substituting for h from equation (i) we have
gt2/8 = gt12/2
This gives t1 = t/2
The total time time taken by the ball (from the instant of projection) to reach a height h/4 while returning is t + t1 = t + (t/2) = 3t/2 = 1.5 t.

(4) The velocity-time graph of an object moving along the x-direction is shown in the figure. What is the displacement of the object when it moves with the maximum acceleration?
(a) 12 m
(b) 8 m
(c) 6 m
(d) 4.8 m
(e) 4.2 m
The object has maximum acceleration from 10 sec to 12 sec since the slope of the velocity-time graph is maximum during this interval. The displacement is given by the area under the velocity-time graph for the interval from 10 sec to 12 sec which is 8 m.
[The displacement of the object when it moves with the maximum acceleration can be calculated using the equation,  s = ut + ½ gt2 as well:
As is evident from the velocity-time graph, the maximum acceleration a is given by
a = Change of velocity/Time = (5 – 3)/(12 – 10) = 1 ms–2
Since the initial velocity u = 3 ms–1 and the time interval t = 2 s, we have
s = (3×2) + (½)×1×22 = 8 m].

(5) A particle moving along the x-axis is initially at the origin with velocity 2 ms–1. If the acceleration a of the particle is given by a = 6t, the position of the particle after 4 seconds is
(a) 24 m
(b) 48 m
(c) 72 m
(d) 96 m
(e) 120 m
The velocity v of the particle is given by
v = adt = ∫6t dt = 3t2 + C where C is the constant of integration which we can find from the initial condition.
Initially (at time t = 0) the particle has velocity 2 ms–1. Therefore from the above expression for velocity we have
C = 2 ms–1
Thus the expression for velocity becomes
v = 3t2 + 2
The position x of the particle is given by
x = v dt = ∫(3t2 + 2)dt = t3 + 2t + C’ where C’ is the constant of integration in this case.
Initially (at t = 0) since the particle is at the origin (where x = 0) we obtain (from the above expression for v)
C’ = 0
Therefore, the expression for the position x becomes
x = t3 + 2t
The position x’ at t = 4 seconds is given by
x’ = 43 + (2×4) = 72 m

You can find a few more useful questions (with solution) in this section here.

## Thursday, November 8, 2012

### AP Physics C - Multiple Choice Practice Questions on Moment of Inertia

"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."
Albert Einstein

As you know, moment of inertia in rotational motion takes the role of mass in translational motion. Even though mass of an object is a constant in Newtonian mechanics, its moment of inertia is not a constant since it depends on the axis of rotation and the position of its constituent particles relative to the axis. Today we shall discuss a few multiple choice practice questions on moment of inertia. Many questions in this section were discussed on this site earlier. You can access them by clicking on the label ‘moment of inertia’ below this post or by trying a search for ‘moment of inertia’ using the search box provide on this page.

Here are the questions:
(1) Moment of inertia in mechanical systems is comparable to one of the following in electrical systems. Pick it out:
(a) Resistance
(b) Magnetic moment
(c) Time constant
(d) Capacitance
(e) Inductance
Because of moment of inertia a mechanical system opposes any change in its state of rotation. Because of inductance an electrical system opposes any change in the current flowing in it. Therefore moment of inertia in mechanical systems is comparable to inductance in electrical systems.
(2) The rotational kinetic energy of a wheel is 1600π2 J. If the wheel is rotating at 20 rev/s, the moment of inertia of the wheel is
(a) 1 kgm2
(b) 2 kgm2
(c) 3 kgm2
(d) π2 kgm2
(e) 4π2 kgm2
The rotational kinetic energy E is given by
E = (½) Iω2 where I is the moment of inertia and ω is the angular velocity (in radian per second). Since the wheel is rotating at 20 rev/s, its angular velocity ω in radian/second is 2π×20 = 40π.
Therefore we have
1600π2 = (½) I×(40π)2 from which I = 2 kgm2
(3) A thin circular disc has moment of inertia I about an axis passing through its centre and perpendicular to its plane. What is its moment of inertia about an axis tangential to its edge and parallel to its plane?
(a) I/4
(b) I/2
(c) 2I/3
(d) 3I/2
(e) 5I/2

Since the moment of inertia of the circular disc about an axis passing through its centre and perpendicular to its plane is I, the moment of inertia I1 about a diameter is I/2 in accordance with the theorem of perpendicular axes. The moment of inertia I2 of the disc about an axis tangential to its edge and parallel to its plane is I1 + MR2 in accordance with the theorem of parallel axes:
I2 = I1 + MR2 = (MR2/4) + MR2
Or, I2 = 5MR2/4
We know that I = MR2/2
Therefore, I2 = 5I/2
The adjoing figure will make things very clear for you.
(4) The radius of gyration of a thin circular ring about a diameter is Rg. The diameter of the ring is
(a) 2 Rg
(b) 4 Rg
(c) √(2Rg)
(d) √(8Rg)
(e) Rg/2
The moment of inertia of a circular ring about a central axis perpendicular to the plane of the ring is MR2 where M  is its mass and R is its radius. The moment of inertia of the ring about a diameter is MR2/2 in accordance with the theorem of perpendicular axes. Since the radius of gyration of the ring about a diameter is Rg we have
MR2/2 = MRg
Therefore R = √(2Rg)
The diameter of the ring = 2R = 2√(2Rg) = √(8Rg)
(5) The moment of inertia of a thin uniform rod about an axis perpendicular to the rod and passing through its centre is I. Moment of inertia of the rod about a parallel axis through its end is
(a) 3I/2
(b) 2I
(c) 5I/2
(d) 3I
(e) 4I
If the mass of the rod is M an the length is L, the moment of inertia I of the rod about an axis perpendicular to the rod and passing through its centre is given by
I = ML2/12
Moment of inertia I1 of the rod about a parallel axis through its end is given by
I1 = (ML2/12) + M(L/2)2, in accordance with the theorem of parallel axes.
Or, I1 = ML2/3 = 4I

Now, see these questions (with solution) involving moment of inertia.