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 evs the unique world of insects and animals sleeping time animal senses sense organs in animals sense organs in humans

The stato-acoustic receptor responds to changes in the ..................

  1. Light and pressure

  2. Pressure and touch

  3. Pain and pressure

  4. Sound and equilibrium.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Statio-acoustic receptors are meant for both sensory functions, i.e., hearing (sound) and maintenance of body balance (equilibrium) of the body. These receptors generate stimuli in response to sound and gravity. The function of  stato-acoustic receptor is hearing and balancing.


Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A certain strings will resonate to several frequencies , the lowest of which is $200$cps.what are the next three higher frequencies to which it resonates? 

  1. $400,600,800$
  2. $300,400,500$
  3. $100,150,200$
  4. $200,250,300$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,  The Lowest frequency is $200cps$

Let  $f$ resonant the fundamental frequency, then the next higher frequency is: $2f,3f,4f$

$2\times200=400cps,3\times200=600,4\times200=800cps$


Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

The string of a violin emits a note of 205 Hz at its correct tension. The string is tightened slightly and then it produces six beats in two seconds with a tuning fork of frequency 205 Hz. The frequency of the note emitted by the taut string is

  1. 211 HZ

  2. 199 Hz

  3. 208 Hz

  4. 202 Hz
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The initial frequency is 205 Hz. When tightened, the string produces 6 beats in 2 seconds, meaning the beat frequency is 6 / 2 = 3 Hz. Tightening a string increases its tension and thus its frequency. Therefore, the new frequency must be 205 + 3 = 208 Hz.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

If $n,2n,3n$ are the fundamental frequencies of the three segments into which a string is divided by placing required number of bridges below it. If $n _0$ is the fundamental frequency of the string, then 

  1. $n _0=3n$
  2. $n _0=6n$
  3. $n _0=\dfrac{3n}{5}$
  4. $n _0=\dfrac{6n}{11}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The fundamental frequency of a string is f = v / (2L). When divided into segments of lengths L1, L2, L3, the frequencies are f1 = v / (2L1) = n, f2 = v / (2L2) = 2n, f3 = v / (2L3) = 3n. The total length L = L1 + L2 + L3 = v/(2n) + v/(4n) + v/(6n) = (6+3+2)v / 12n = 11v / 12n. The fundamental frequency of the whole string is f0 = v / (2L) = v / (2 * 11v / 12n) = 6n / 11.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

Two tuning forks when sounded together produce 5 beat per second. The first tuning fork is in resonance with 16.0 cm wire of a sonometer and the second is in resonace with 16.2 cm wire of the same sonometer. The frequencies of the tuning forks are

  1. 100 Hz,105 Hz

  2. 20 Hz,205 Hz

  3. 300 Hz,305 Hz

  4. 400 Hz,405 Hz
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The frequency of a sonometer wire is inversely proportional to its length, f proportional to 1/L. Thus, f1 * L1 = f2 * L2, meaning f1 * 16.0 = f2 * 16.2. Also, the beat frequency is f2 - f1 = 5. Solving these simultaneous equations: 16.0 f1 = (f1 + 5) * 16.2, giving 16.0 f1 = 16.2 f1 + 81, so 0.2 f1 = 81, leading to f1 = 400 Hz and f2 = 405 Hz.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A sonometre wire resonates with a given tuning forck forming standing waves with five antinodes between the two bridges when a mass of $9kg$is suspended from the wire. When this mass is replaced by mass $M$, the wire resonates with the same positions of the bridges. Then find the value of square roof of $M$.

  1. $5$
  2. $10$
  3. $25$
  4. $None$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The frequency of vibration of a string 

$n=\dfrac{p}{2l}\sqrt{\dfrac{T}{m}}$
Also number of loops = Number of antinodes.
Hence with 5 antinodes and hanging mass of 9 kg. we have p=5 and T=9g
So,
$n _1=\dfrac{5}{2l}\sqrt{\dfrac{9g}{m}}$
With 3 antinodes and hanging mass M we have p=3 and T=Mg so,
$n _2=\dfrac{3}{2l}\sqrt{\dfrac{Mg}{m}}$
$\because n _1=n _2$
$\dfrac{5}{2l}\sqrt{\dfrac{9g}{m}}=\dfrac{3}{2l}\sqrt{\dfrac{Mg}{m}}$
Squaring both side we get
$25\times9=9\times M$
$M=25\ kg$

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A $12m$ long vibrating string has the speed of wave $48 m/s$ to what frequency it will resonate?

  1. $2cps$
  2. $4cps$
  3. $6cps$

  4. All of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given, $S=48m/s,l=12m$

So, Equation of the fundamental frequency:

$v\dfrac{v}{2l}=\dfrac{48}{2\times12}=2cps$

The string will resonate at fundamental frequency as well as first overtone second overtone and so on.
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

An open tube is in resonance with string (frequency of vibration of tube in $n _{0}$. If tube is dipped on water is that 75% of length of tube is inside water, then the ratio of the frequency of tube to string now will be 

  1. 1

  2. 2
  3. $\dfrac{2}{3}$
  4. $\dfrac{3}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For open tube no $ = \dfrac{V}{2l} $
For closed tube length available for resonance
$ l^{1} ,l\times \dfrac{25}{100} = \frac{l}{4} $
fundamental frequency of water filled tube 
$ n _{1}\dfrac{V}{4l^{1}} = \frac{V}{4(l/4)} $ $(\because l^{1}= l/4) $
$ \therefore \dfrac{V}{l} = 2n _{0} = 1 $
$ = \dfrac{n}{n _{0}} = 2 $
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A string is properly tuned:

  1. When the beat frequency vanishes.

  2. When the beat frequency is maximum.

  3. When the beat frequency is minimum.

  4. When the beat frequency is between maximum and minimum.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Beats are produced when two sound waves of slightly different frequencies interfere. When the two frequencies are identical, the beat frequency becomes zero, meaning the beat frequency vanishes, indicating that the string is properly tuned.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A man generates a ssmmetrical pulse in a string by moving his hand up and down. At $t = 0$ the how hand mowes downuard: The pulse travels with speed of 3$\mathrm { m } / \mathrm { s }$ on the string $&$ his hands passe 6 in each secand from the mean position. Then the point on the string at a distance 3$\mathrm { m }$ will reach its topper arreme first time at time t=

  1. 0.25 sec.

  2. 1 sec
  3. $\frac { 13 } { 12 } \mathrm { sec }$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Hand completes 6 oscillations per second, so period T = 1/6 s. At t=0, hand moves downward. Pulse travels at 3 m/s. Point at 3 m reaches upper extreme when pulse arrives and phase is maximum. Time = distance/speed + T/4 = 3/3 + (1/6)/4 = 1 + 1/24 = 25/24 ≈ 0.25 s.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

1 meter long stretched wire of a sonometer vibrates with its fundamental frequency of 256 Hz. If the length of the wire is decreased to 25 cm and the tension remains the same, then the fundamental frequency of vibration will be:-

  1. 64 Hz

  2. 256 Hz

  3. 512 Hz

  4. 1024 Hz
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The fundamental frequency of a stretched wire is inversely proportional to its length, f proportional to 1 / L, when tension and mass density remain constant. When the length is decreased from 1 m (100 cm) to 25 cm, the length is reduced by a factor of 4, so the fundamental frequency increases by a factor of 4. Thus, f2 = 256 * 4 = 1024 Hz.