Waves and sound-05-OBJECTIVE SOLUTION (LEVEL-I)
OBJECTIVE SOLUTION (LEVEL-I)
1. Determinant of equation must be equal to zero for single root thus
Ans. (c)
2. Ans (d)
3.
hence
Ans. (a)
4.
.
Ans. (b)
5. .
Ans. (b)
6. Speed of sound depends on atmospheric conditions.
Ans. (a)
7. = 36% .
Ans. (b)
8.
.
Ans. (d)
9. Intensity varies as so answer will be (d) only
Ans. (d)
10. &
m.
Ans. (c)
11.
.
Ans. (d)
12.
.
Ans. (b)
13.
Ans. (d)
14.
.
Ans. (a)
15. & Hz
cm/s.
Ans. (a)
OBJECTIVE SOLUTION (LEVEL-II)
1. should be zero for a maximum / minimum
where .
Ans. (b)
2. leads since it is +ive valued when is zero.
lags since it is –ive valued when is zero.
Ans. (a)
3.
sec.
= 200 m/s.
Ans. (c)
4. 5 = 10
.
Ans. (a)
5.
Ans. (b)
6.
Ans. (a)
7. ,
.
Ans. (a)
8. Since beats is increased then
.
Ans. (c)
9. Since net amplitude is zero, so energy will be purely kinetic.
Ans. (b)
10. Ans. (a)
11.
= 240 Hz.
12. Path difference when the waves are in phase
.
Ans. (c)
13.
.
Since
hence = 0.072 m.
Ans. (b)
14. And (b)
15.
,
.
SUBJECTIVE SOLUTION (LEVEL-I)
1.
.
2. (a) mm.
(b) m.
(c) sec., Hz.
3.
.
4. Rod behaves as a closed organ pipe
.
5.
.
If both ends are open
No.
6.
Minimum value of would be 1.
hence
m/s.
7. … (i)
At tension increased the frequency of , beats will decreases as given. ____ equation (i) must be in the form of
Hz.
8.
Hz.
9. (a) Hz
(b) similarly
= 388.3 Hz.
10. (i) Travelling, and
m/s
(ii) cm
(iii) Hz
(iv) m.
11.
cm
Now phase difference path difference
(i) rad.
rad.
Similarly (ii) rad.
(iii) rad
(iv) rad.
12. rad, rad
cm
(i) cm/s
cm/s
(ii) cm/s2
cm/s2.
13. The particle displacements are given by
where a = 0.1 m.
We are given - 0.05 = 0.1 sin …(i)
and + 0.05 = 0.1 sin …(ii)
Eqs. (i) and (ii) give
and
Subtracting these equations, we get m. Now . Therefore, time taken is
.
14. Let be the frequency of the source and that of the th harmonic of a closed pipe. The source will resonantly excite that harmonic mode of the pipe for which
for any value of p = 1, 2, 3, …… Now for a closed pipe, we know that
Therefore, for resonance,
.
15. Since the source of sound is stationary, the wavelength remains unchanged by the motion of the observer. Therefore,
m
The frequency of sound heard by the observer is
Hz.
SUBJECTIVE SOLUTION (LEVEL-II)
1. On loading frequency will decrease and also beats increases thus
Hz.
2.
.
3. cm
.
and
cm
cm
cm’s.
4. As calculated by , the velocity of sound is equal thus the only possible figure so that the standing waves will be developed in the ‘system’ is
Thus min frequency = 180 Hz.
Alternative :
Since & are integers thus .
.
5. and
since for minimum is 3 is integer number)
Hence
.
6. (a) 500 Hz, since no relative motion
(b) Hz.
(c) 500 Hz
Hz
where Velocity of observer w.r.t. to medium
m/s
and m/s
Hz
Hz.
7. Let
where
so by principle of superposition
where
for to be max or min
i.e.,
either or,
maximum at
;
Intensity is maximum (with two different values) i.e. within the second we got two maxi and two minima so beat frequency is 2.
8. Observer will listen max./min. frequency when source velocity is pointed at observer.
.
.
9.
m.
m/s.
Apparent
.
m/s
Hz.
10. Acceleration
Now,
where
etc.
SUBJECTIVE SOLUTION (LEVEL-III)
1. (a)
(b)
(c)
.
(d) and
and .
2.
… (i)
and … (ii)
from (i) and (ii)
hence
(from (i))
.
3.
as tension decrease, it will decrease the frequency, , so beats will be decreased thus
.
4.
.
.
5. Path difference,
(a) For minimum intensity
.
(b) For maximum intensity
(c)
when
so 2 maximum.
6.
7. (i) The frequency heard after reflection will be the same as heard directly from the source so beat frequency will be zero in this case.
(ii)
beats.
(iii)
beats.
8.
and
.
9. (a)
loops.
(b)
.
(c)
where
(d)
.
10.
where
.
.
1. Frequency received in case I
and in case II
Obviously
Beat frequency
or, .
2. Frequency of second overtone of the closed pipe
(Given)
.
Substituting speed of sound in air = 330 m/s
.
(b) Open end is displacement antinode. Therefore, it would be a pressure node
or,
Pressure amplitude at , can be written as
Therefore, pressure amplitude at or (15/32) m will be
.
(c) Open end is a pressure nod i.e.,
Hence Mean pressure
(d) Closed end is a displacement node or pressure antipode.
Therefore,
and .
3. Tension
Amplitude of incident wave
Mass per unit length of wire is
and mass per unit length of wire is
(a) Speed of wave in wire is
and speed of wave in wire is
Time taken by the wave pulse to reach from to is –
.
(b) The expressions for reflected and transmitted amplitudes and in terms of and are as follows :
and
Substituting the values we get
i.e. the amplitude of reflected wave will be 1.5 cm. Negative sign of indicates that there will be a phase of in reflected wave.
Similarly
i.e. the amplitude of transmitted wave will be 2.0 cm.
4. Velocity of sound in water is
Frequency of sound in water will be
.
(a) Frequency of sound detected by receiver (observer) at rest would be ¬–
.
(b) Velocity of sound in air is
= 344 m/s.
Frequency does not depend on the medium. Therefore, frequency in air is also .
Frequency of sound detected by receiver (observer) in air would be
.
5. (a) Frequency of second harmonic in pipe frequency
of third harmonic in pipe
or, (as )
or,
or, (as )
or, .
(b) Ratio of fundamental frequency in pipe and in pipe is:
(as )
(as )
substituting from part (a), we get
.
6. Fundamental frequency,
speed of sound
or, m/s.
7 or,
The amplitude at a distance from is given by
Total mechanical energy at of length is
or, … (i)
Here and .
Substituting these values in equation (i) and integrating it from to , we get total energy of string,
.
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