Physics · Science General

Acoustics and Sound Waves

2,160 Questions

Acoustics and sound waves deal with mechanical vibrations traveling through media like air and water. Key concepts include wave reflection, beats, echoes, the Mach number, and infrasound. These physics fundamentals are regularly tested in general science sections of multiple competitive exams.

Sound wave propagationEchoes and reflectionWave interferenceMach numberInfrasound frequency

Acoustics and Sound Waves Questions

Multiple choice biology sensory organs of humans - sense organs ear : the organ of hearing and equilibrium ear and its function ear

Following is a list of the events (in a random order) that lead to the formation of an auditory impulse.
(i) Vibration is transferred from the malleus to the incus to the stapes.
(ii) Basilar membrane moves up and down.
(iii) Nerve impulse is transmitted in cochlear nerve to auditory cortex of brain for impulse analysis and recognitions.
(iv) Sound waves pass through ear canal.
(v) Stereocilia of hair cells of organ of Corti rub against tectorial membrane.
(vi) Sound waves cause ear drum to vibrate.
(vii) Nerve impulse is generated.
(viii) Vibrations move from fluid of vestibular canal to the fluid of tympanic canal.
(viii) Vibrations move from fluid of vestibular canal to the fluid of tympanic canal.
(ix) Membrane at oval window vibrates.
Which of the following options represents these events in a correct order?

  1. (iv), (vi), (i), (ix), (viii), (ii), (v), (vii), (iii)

  2. (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix)

  3. (ix), (viii), (vii), (vi), (v), (iv), (iii), (ii), (i)

  4. (iv), (vi), (i), (viii), (ix), (ii), (v), (vii), (iii)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ear not only detects sound but also notes its direction, judges its loudness and determines is pitch.Auditory impulse form through different steps:

1. Sound waves are collected by the pinna and directed inward through external auditory canal.
2. Here they strike the tympanic membrane. The latter begins to vibrate at the same frequency as that of the sound waves.
3. From tympanic membrane, the vibration transmitted across the tympanic cavity by the ear ossicles.The outer ossicles called malleus, inner called stapes, middle ossicles called incus.
4. Stapes is fit into oval window which vibrates. The ear ossicles transmit the vibrations from tympanic membrane to the internal ear and also amplify them 20 times.
5.  Increase in force is important because the sound waves are transmitted from air to a fluid medium.
6. From here, the vibrations are transferred to the basilar membrane and the perilymph in the scala tympani, and are finally dissipated into the air of the middle ear as vibrations of the round window membrane.
7. Vibrations of the floor make the "sensory hair" of receptor cells in the organ of corti move in the overlying gelatinous membrane, and get distorted.
8. This stimulation causes  depolarisation of the receptor cells and initiation of receptor( action) potential in the fibres of the auditory nerve.
9. The latter carries the impulses to the cerebral cortex, which interprets the impulses as sound.
So, the correct answer is '(iv), (vi),(i), (ix), (viii), (ii), (v), (vii), (iii)'.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The frequency of vibration of a sonometer wire is directly proportional to linear density of the wire:

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt{(T/\mu)}; \mu $ is the linear density of the material of the wire
 
Thus, frequency of vibration is inversely proportional to $\sqrt(\mu)$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A knife edge divides a sonometer wire in two parts which differ in length by 2 mm. The whole length of the wire is 1 meter. The two parts of the string when sounded together produce one beat per second. Then the frequency of the smaller and longer pans.in Hz,are

  1. 250.5 and 249.5

  2. 249.5 and 250.5

  3. 124.5 and 125.5

  4. 125.5 and 124.5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the frequency formula f = (1/2L) * sqrt(T/m), the difference in frequencies for two segments of lengths L1 and L2 is 1 Hz. Given L1 + L2 = 1m and L1 - L2 = 0.002m, we find L1 = 0.501m and L2 = 0.499m. The frequencies are proportional to 1/L, leading to 124.5 Hz and 125.5 Hz.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A sonometer wire of length $l _1$ vibrates with a frequency 250 Hz. If the length of wire is increased then 2 beats/s are heard. What is ratio of the lengths of the wire?

  1. 124 : 125

  2. 250 : 313

  3. 5 : 3

  4. 41 : 57
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The frequency of sonometer wire is given by
$n=\dfrac{p}{2l}\sqrt{(\dfrac{T}{m})}$
or $n\propto \dfrac{1}{l}$     ...(i)
$\therefore \dfrac{n _1}{n _2}=\dfrac{l _2}{l _1}$
or $\dfrac{250}{250-2}=\dfrac{l _2}{l _1}$
or $\dfrac{l _1}{l _2}=\dfrac{248}{250}$
$=\dfrac{124}{125}=124: 125$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The tension in the sonometer wire is decreased by 4% by loosening the screws. It fundamental frequency

  1. remains same

  2. increases by 2%

  3. decreases by 2%

  4. frequency becomes imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt(T/\mu)$

The fractional change in frequency with change in tension is given by $\Delta f/f = -(\Delta T/2T)$.

Thus, the frequency decreases by 2%

The correct option is (c)

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

Fundamental frequency of a sonometer wire is n. If the length and diameter of the wire are doubled keeping the tension same, then the new fundamental frequency is :

  1. $\dfrac{n}{2\sqrt{2}}$
  2. $\sqrt{2}n$
  3. $\dfrac{n}{4}$
  4. $\dfrac{2n}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Frequency of vibration of sonometer wire is given as
$n=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}=\dfrac{1}{2l}\sqrt{\dfrac{T}{\pi r^2d}}\Rightarrow n\propto \dfrac{1}{\sqrt{(d)}}$
If the length and diameter of the wire are doubled. The new frequency will be $\dfrac{n}{4}$.
Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A wire with linear density of 3 gm/mm is used as a sonometer wire for producing vibrations of frequency 50 Hz. This length of this wire is now halved, while the tension is reduced by 1/4th of the initial tension. What will be the frequency of vibrations produced:

  1. 10 Hz

  2. 30 Hz

  3. 50 Hz

  4. 70 Hz
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt(T/\mu)$

since length is halved and tension is made 1/4th the initial tension, their ratio $\sqrt(T)/L$ remains constant. Thus, frequency now depends only on linear mass density

Since the same wire is used, linear mass density dosen't change and hence the frequency also remains same

The correct option is (c)

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

Two identical sonometer wires have a fundamental frequency of $500$ Hz, when kept under the same tension. What fractional increase in the tension of one wire would cause an occurrence of $5$ beats/sec, when both wires vibrate together?

  1. 2

  2. 3

  3. 4

  4. 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$n\propto V$
and $V\propto \sqrt{T}$
$\Rightarrow n\propto \sqrt{T}$ ..$(1)$
$5$ beats/sec are obtained when the frequency of one become $505$ Hz i.e. percentage increase in frequency is $1\%$
From $(1)$ Percentage increase in $\eta =1\%$
$\Rightarrow \%$ increases in tension $=2\%$
(Note that method is applicable for small changes only).

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

When the length of the vibrating segment of a sonometer wire is increased by 1%, the percentage change in its frequency is

  1. 100/101

  2. 99/100

  3. 1

  4. 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt(T/\mu)$

The fractional change in frequency with change in length is given by $\Delta f/f = -\Delta L/L$.

Thus, the frequency decreases by 1%

The correct option is (c)

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A brick is hung from a sonometer  wire. If the brick is immersed in oil, then frequency of the wire will 

  1. increase due to buoyancy

  2. decrease

  3. remains unchanged

  4. increase due to viscosity of oil
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When immersed in oil , the tension in the string decreases due to force of buoyancy.
$\therefore$ as $f\  \alpha\  \sqrt{T}$
frequency decreases  

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A sonometer wire of length 114 cm is fixed at the both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio 1:3:4?

  1. at 36 cm and 84 cm from one end

  2. at 24 cmand 72 cm from one end

  3. at 48 cm and 96 cm from one end

  4. at 72 cm and 96 cm from one end
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Frequencies are in ratio 1:3:4, so lengths must be in ratio 1/1 : 1/3 : 1/4, which is 12:4:3. Total parts = 19. Lengths are (12/19)*114 = 72cm, (4/19)*114 = 24cm, and (3/19)*114 = 18cm. Placing bridges at 72cm and 96cm (72+24) creates segments of 72, 24, and 18 cm.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

If the length of the wire of a sonometer is halved the value of resonant frequency will get:

  1. doubled

  2. halved

  3. four times

  4. eight times
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ f=\dfrac{1}{2e}\sqrt{\dfrac{t}{\mu }}$
$f\alpha \dfrac{1}{l}$
$\therefore \dfrac{f _{1}}{f _{2}}=\dfrac{l _{2}}{l _{2}}=\dfrac{1}{2}$
$\Rightarrow f _{2} = 2f _{1}$
$\therefore $ frequency is doubled