Tag: sound

Questions Related to sound

 769Hz longitudinal wave in air has a speed of 344m/s. At a particular instant, what is the phase difference (in degrees) between two points 5.0 cm apart?

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. 30

  2. 40

  3. 45

  4. 60

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Correct Option: B
Explanation:

Given that,

The frequency of wave is $\nu = 769 Hz$

The speed of wave is $v = 344 ms^{-1}$

The wavelength of wave is given by

$\lambda = \dfrac{v}{\nu}$

$\lambda = \dfrac{344}{769}$

$\lambda = 0.447 m = 44.7 cm$

Hence, the phase difference between two points 5.0 cm apart is

$\dfrac{5}{44.7} \times 360^\circ = 0.1118 \times 360^\circ = 40.24^\circ \approx40^\circ$

A rod $70\space cm$ long is clamped from middle. The velocity of sound in the material of the rod is $3500\space ms^{-1}$. The frequency of fundamental note produced by it is :

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. $3500\space Hz$

  2. $2500\space Hz$

  3. $1250\space Hz$

  4. $700\space Hz$

Show answer
Correct Option: B
Explanation:

SInce rod is clamped at middle, therefore only nodes can form at that point, and at free end only antinode can form, therefore for fundamental mode of frequency,

$\lambda/4=L/2$

$\lambda=2L=2\times 70 cm=1.4 m$

$v=3500\  ms^{-1}$ is given, 

We know, $v=f \times \lambda$ or $f = \dfrac{v}{\lambda}=3500/1.4=2500 Hz$

Option "B" is correct.

A sonometer wire, $100\ \text{cm}$ in length has a fundamental frequency of $330\ \text{Hz}$. The velocity of propagation of transverse waves along this wire is :

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. $330\ \text{ms}^{-1}$

  2. $660\ \text{ms}^{-1}$

  3. $115\ \text{ms}^{-1}$

  4. $990\ \text{ms}^{-1}$

Show answer
Correct Option: B
Explanation:

For the fundamental frequency $(f _0=330 Hz),\ \lambda=2 L=200\ \text{cm} = 2\ \text{m}$


then from the formula, $v=f _0 \times \lambda= 330 \times 2.0= 660 \text{ms}^{-1}$

If the frequency of a sound wave is increased by 25%, then the change in its wavelength will be

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. 25% decrease

  2. 20% decrease

  3. 20% increase

  4. 25% increase

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Correct Option: B
Explanation:
The frequency  of wave is given by

$\nu = \dfrac{v}{\lambda}$

$\lambda = \dfrac{v}{\nu}$

When the frequency of a sound wave is increased by 25%, then the new 
wavelength is

$\lambda' = \dfrac{v}{\nu+\dfrac{25}{100}\nu}$

$\lambda' = \dfrac{v}{\nu+\dfrac{1}{4}\nu}$

$\lambda' = \dfrac{v}{\dfrac{5}{4}\nu}$

$\lambda' = \dfrac{4v}{5\nu}$

Hence, the percent change in wavelength is

$\dfrac{\lambda - \lambda'}{\lambda} \times 100 = \dfrac{\dfrac{v}{\nu} - \dfrac{4v}{5\nu}}{\dfrac{v}{\nu}} \times 100$

$\Rightarrow -\dfrac{1}{4} \times 100 = -20$%.

Hence, wavelength decreases by 20%

Sound waves of wavelength $\lambda $ travelling with velocity $v$ in a medium enter into another medium in which their velocity is $4v$. The wavelength in $2^{nd}$ medium is :

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. $4\lambda$

  2. $\lambda $

  3. $\lambda/4 $

  4. $ 16\lambda $

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Correct Option: A
Explanation:

From $v=n\lambda $ we find $\lambda\propto v$ because frequency n is constant .

Therefore ,
new wavelength = $4\lambda $

A $40\ cm$ long brass rod is dropped, one end first on to a hard floor but it is caught before it topples over. With an oscilloscope it is determined that the impact produces a $3\ kHz$ tone. The speed of sound in brass is:

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. $1200\ m/s$

  2. $2400\ m/s$

  3. $3600\ m/s$

  4. $3000\ m/s$

Show answer
Correct Option: B
Explanation:

Both ends are free and therefore antinodes are formed.
The relation between the wavelength of the wave and the length of the rod for fundamental frequency will be:

$\Rightarrow l=\dfrac{\lambda}{2} \ \ \Rightarrow \lambda=2l$

The speed of the wave in the rod is:
$v=f\lambda = 2\times 40\times 3\times 10^{3}$$=2400\ ms^{-1}$

The frequency of a man's voice is 300 Hz and its wavelength is 1 meter. If the wavelength of a child's voice is 1.5 m, then the frequency of the child's voice is :

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. 200 Hz

  2. 150 Hz

  3. 100 Hz

  4. 350 Hz.

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Correct Option: A
Explanation:

$\nu _1\lambda _1= \nu _1\lambda _1$ since $v= \nu\lambda$ is same for both a man and child.

$ \therefore 300 \times 1 =  \nu _2 \times  1.5$


$ \Rightarrow \nu _2 = 200 : Hz$

Sound wave are not polarized because:

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. Their speed is less

  2. A medium is needed for their propagation

  3. These are longitudinal

  4. Their speed depends on temperature

Show answer
Correct Option: A
Explanation:
   Sound wave are not polarized because,their speed is less.So the option is A.

769Hz longitudinal wave in air has a speed of 344m/s. How much time is required for the phase to change by 90 degrees at a given point in space? 

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. 0.325 ms

  2. 3.25 ms

  3. 32.5 ms

  4. 325 ms

Show answer
Correct Option: A
Explanation:

Given that, 

The frequency of wave is $\nu = 769 Hz$.

The speed of wave is $v = 344 ms^{-1}$.

We know, the frequency of a wave is the number of oscillations per 
unit time. The period (T) of a wave is the time that it takes for one complete oscillation, i.e., to change its phase through $360^\circ$.
Hence,

$T = \dfrac{1}{\nu}$

$T = \dfrac{1}{769}$

$T = 0.0013 s$

$T = 1.3 ms$
Here, the phase change is 90 degrees at a given point in space.  

$\dfrac{90}{360} = \dfrac{1}{4}$

Hence, the time period required to complete one fourth oscillation is

$T = \dfrac{1.3}{4} = 0.325 ms$

The musical note A is a sound wave. The note has a frequency of 440 Hz and a wavelength of 0.784 m. Calculate the speed of the musical note in m/s.

physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

  1. 343

  2. 345

  3. 34

  4. 346

Show answer
Correct Option: B
Explanation:

Let,

The frequency of the musical note is $\nu = 440 Hz = 440 s^{-1}$

The wavelength of the musical note is $\lambda = 0.784 m $

Now, the frequency of the musical note is 
$\nu = \dfrac{v}{\lambda}$

where, v is the speed of the musical note 
Hence, the speed of the musical note is
$v = \nu \lambda$

$v = 440 \times 0.784$

$v = 344.96 \approx 345 ms^{-1}$