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 physics oscillations and waves standing waves in strings standing waves reflection of waves

In an organ pipe(may be closed or open) of $99$ cm length standing wave is setup, whose equation is given by longitudinal displacement.
$\xi =(0.1mm)\cos \dfrac{2\pi}{0.8}(y+1cm)\cos 2\pi (400)t$
where y is measured from the top of the tube in meters and t is second. Here $1$cm is the end correction.
The air column is vibrating in :

  1. First overtone

  2. Fifth harmonic

  3. Third harmonic

  4. Fundamental mode
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is of the form cos(k(y+c))cos(wt). The angular frequency w = 2*pi*400, so f = 400 Hz. The wave number k = 2*pi/0.8, so wavelength lambda = 0.8 m = 80 cm. With end correction, the effective length is 99 + 1 = 100 cm. For a closed pipe, L = (2n-1)lambda/4. 100 = (2n-1)80/4 = (2n-1)20. 5 = 2n-1, so n=3. This corresponds to the 5th harmonic.

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

In a reasonance tube experiment, a closed organ pipe of lenght $120$ cm is used. initially it is completely fiiled with water. It is vibrated with tuning fork of frequency $340$ Hz. To achieve reasonance the water level is lowered then (given ${V _{air}} = 340m/\sec $., neglect end correction):

  1. minimum lenght of water column to have the resonance is 45 cm.

  2. the distance between two successive nodes is 50 cm.

  3. the maximum lenght of water column to resonance is 95 cm.

  4. the distance between two successive nodes is 25 cm.
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

The frequency of A note is $4$ times that of B note. The energies of two notes are equal. The amplitude of B note as compared to that of A note will be:

  1. double

  2. equal

  3. four times

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

$E$ is for energy, $A$ for amplitude and $f $ for frequency.

As per the problem $E _{A} = E _{B}$
Hence, $f _{A} \times A _{A}^{2} = f _{B} \times A _B^2$
 $4f _{B} \times A _{A}^{2} = f _{B} \times A _B^2$
Hence, $2A _A = A _B $

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

A string vibrates in 5 segment to a frequency of 480 Hz. The frequency that will cause it to vibrate in 2 segments will be

  1. 96 Hz

  2. 192 Hz

  3. 1200 Hz

  4. 2400 Hz
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

5 segments implies $\lambda = \dfrac{2}{5}l$
$\nu = \dfrac{v}{\lambda} = \dfrac{5v}{2l} = 480Hz$
If the string is in 2 segments.
$\lambda = l$
$\nu = \dfrac{v}{\lambda} = \dfrac{2}{5} \dfrac{5v}{2l} = \dfrac{2}{5} 480 = 192Hz$
Hence option B is correct.

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

A pipe of length $l _1$ closed at one end is kept in a chamber of gas density $1$. A second pipe open at both ends is placed in the second chamber of gas density $2$. The compressibility of both the gases is equal.Calculate the length of the second pipe if the frequency of the first overtone in both the cases is equal.

  1. $\displaystyle \dfrac{4}{3}l _{1}\sqrt{\dfrac{\mathrm{p} _{2}}{\mathrm{p} _{1}}}$
  2. $\displaystyle \dfrac{4}{3}l _{1}\sqrt{\dfrac{\mathrm{p} _{1}}{\mathrm{p} _{2}}}$
  3. $l _{1}\sqrt{\dfrac{\mathrm{p} _{2}}{\mathrm{p} _{1}}}$
  4. $l _{1}\sqrt{\dfrac{\mathrm{p} _{1}}{\mathrm{p} _{2}}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$l _{1}=\displaystyle \dfrac{3}{4}\dfrac{\mathrm{v} _{1}}{\mathrm{f} _{1}}$ , $l _{2}=\displaystyle \dfrac{\mathrm{v} _{2}}{\mathrm{f} _{2}}$

$\dfrac{3\mathrm{v} _{1}}{4l _{1}}=\dfrac{\mathrm{v} _{2}}{l _{2}}$

$l _{2}=\displaystyle \dfrac{4l _{1}\mathrm{v} _{2}}{3\mathrm{v} _{1}}=\dfrac{4l _{1}}{3}\sqrt{\dfrac{\mathrm{p} _{1}}{\mathrm{p} _{2}}}$

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe 2 m long. The length of the open pipe is

  1. 8 m

  2. 4 m

  3. 2 m

  4. 1 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$l _0=$ length of organ pipe (open)

$l _c=$ length of close organ pipe
frequency of open pipe$=\cfrac{3V}{2l _0}$
frequency of close pipe $=\cfrac{3V}{2l _c}$
$\therefore \cfrac{3V}{2l _0}=\cfrac{3V}{4l _c}\2l _0=4l _c\2\times2m=4m$
B is the correct option.

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

A guitar string is $90 cm$ long and has a fundamental frequency of $124 Hz$. To produce a fundamental frequency of $186 Hz$, the guitar should be pressed at ?

  1. $60 cm$
  2. $30 cm$
  3. $20 cm$
  4. $ 10 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here, $L _1 = 90 cm, \upsilon _1 = 124 Hz, \upsilon _2 = 186 Hz, L _2=?$

According to the law of length, $\upsilon _2L _2 = \upsilon _1L _1$

$\therefore \upsilon _2= \dfrac{\upsilon _1L _1}{\upsilon _2} = \dfrac{124 \times 90}{186} = 60 cm$

Multiple choice classification of computer introduction to the information age physics

Signal becomes weak with increase in distance travelled because of. 

  1. Modulation

  2. Attenuation

  3. Switching

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

Signal becomes weak with increase in distance travelled because of Attenuation.

Consequently, positive attenuation causes signals to become weaker when travelling through the medium. ... Causes of attenuation in both signal frequency and range between the end points of the medium, affect the amount of signal reduction. As the range increases, attenuation also increases.

In a nutshell, attenuation is the loss of transmission signal strength measured in decibels (dB). As attenuation increases, the more distorted and unintelligible the transmission (e.g. a phone call or email you're trying to send) becomes. Inherent attenuation can be caused by a number of signaling issues including:

Transmission medium - All electrical signals transmitted down electrical conductors cause an electromagnetic field around the transmission. ... Crosstalk from adjacent cabling causes attenuation in copper or other conductive metal cabling.

Attenuation is a general term that refers to any reduction in the strength of a signal. ... Sometimes called loss, attenuation is a natural consequence of signal transmission over long distances. The extent of attenuation is usually expressed in units called decibels (dBs).

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

The intensity of the sound gets reduced by $10$% on passing through a slab. The reduction in  intensity on passing through two consecutive slab, would be 

  1. $20$%
  2. $50$%
  3. $19$%
  4. $5$%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If intensity is reduced by 10 percent, 90 percent (0.9) remains. After two slabs, the remaining intensity is 0.9 * 0.9 = 0.81. The total reduction is 1 - 0.81 = 0.19, which is 19 percent.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Sounds from two identical $S _1$ and $S _2$ reach a point  P. When the sounds reach directly, and in the same phase, the intensity at $P$ is $I _0$. The power of $S _1$ is now reduced by $64\%$ and the phase difference between $S _1$ and $S _2$ is varied continuously. The maximum and minimum  intensities recorded at P are mow $I _{max}$ and $I _{min}$ 

  1. $I _{max}=0.64I _0$
  2. $I _{min}=0.36I _0$
  3. $\dfrac{I _{max}}{I _{min}}=16$
  4. $\dfrac{I _{max}}{I _{min}}=\dfrac{1.64}{0.36}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial intensity I0 = (sqrt(I1) + sqrt(I2))^2. Since S1 and S2 are identical, I1 = I2 = I, so I0 = (2*sqrt(I))^2 = 4I. Power of S1 reduced by 64 percent means new I1 = 0.36I. New Imax = (sqrt(0.36I) + sqrt(I))^2 = (0.6 + 1)^2 * I = 2.56I. Since I0 = 4I, I = I0/4. Thus Imax = 2.56 * (I0/4) = 0.64I0.

Multiple choice physics sound audible, infrasonic and ultrasonic sounds different sounds range of hearing

The sounds having frequency less than 20 Hz are called ultrasonics.

  1. True

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

Human beings can normally hear sounds with a frequency between about 20 Hz and 20,000 Hz. Sounds with frequencies below 20 hertz are called infrasound. Infrasound is too low-pitched for humans to hear.

so the answer is B.