Tag: sound as a wave of disturbance

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Questions Related to sound as a wave of disturbance

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

The frequency of a man's voice is $300\space Hz$. If the velocity of sound waves is $336\space ms^{-1}$, the wavelength of the sound is

  1. $1.12\space m$
  2. $300\times336\space m$
  3. $330/336\space m$

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

We are given, frequency $f=300 Hz$, and velocity $v=336 ms^{-1}$,

From the relation $v=f \times \lambda$,

$\lambda = v/f =336/300=1.12 m$

Option "A" is correct.

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

 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?

  1. 30

  2. 40

  3. 45

  4. 60
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

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 :

  1. $3500\space Hz$
  2. $2500\space Hz$
  3. $1250\space Hz$
  4. $700\space Hz$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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 :

  1. $330\ \text{ms}^{-1}$
  2. $660\ \text{ms}^{-1}$
  3. $115\ \text{ms}^{-1}$
  4. $990\ \text{ms}^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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


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

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

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

  1. 25% decrease

  2. 20% decrease

  3. 20% increase

  4. 25% increase
Reveal answer Fill a bubble to check yourself
B Correct answer
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%
Multiple choice physics sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

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 :

  1. $4\lambda$
  2. $\lambda $
  3. $\lambda/4 $
  4. $ 16\lambda $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Therefore ,
new wavelength = $4\lambda $

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

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:

  1. $1200\ m/s$
  2. $2400\ m/s$
  3. $3600\ m/s$
  4. $3000\ m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
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}$

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

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 :

  1. 200 Hz

  2. 150 Hz

  3. 100 Hz

  4. 350 Hz.
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

Multiple choice physics stationary waves sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

The wave produced in a resonance tube is

  1. Longitudinal

  2. Transverse

  3. Transverse stationary

  4. Longitudinal stationary
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Waves produced in a resonance tube are sound waves. Two sound waves in opposite direction interfere with each other to create resonance. As sound waves are longitudinal waves, waves produced are longitudinal stationary.