Physics · Science General

Acoustics and Sound Waves

2,006 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

Which of the following term is used for sound below 20 Hz?

  1. Infrasound

  2. Ultrasound

  3. Echo

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The human ear can generally hear sounds with frequencies between 20 Hz and 20 kHz. Although hearing requires an intact and functioning auditory portion of the central nervous system as well as a working ear, human deafness most commonly occurs because of abnormalities of the inner ear, rather than in the nerves or tracts of the central auditory system. Sound below 20 Hz is considered infrasonic, which the ear cannot process. Therefore, the correct answer is option A.

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

Sensation of hearing is produced as a result of the

  1. Presence of the helicotrema

  2. Vibrations set in the external auditory meatus

  3. Vibrations of the ear ossicles

  4. Nerve impulses generated by the hair cells of the organ of Corti
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Within the bony labyrinth, is the membranous labyrinth; filled with the fluid called as endolymph. The membranous labyrinth within the cochlea contains the organ of Corti containing the inner and outer hair cells, that generate the nerve impulses required for hearing. Nerve impulses generated in the inner ear travel along the vestibulocochlear nerve (cranial nerve VIII) to the brain.

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