Tag: determining wavelength and speed of sound

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Questions Related to determining wavelength and speed of sound

Multiple choice physics superposition of waves-2: stationary (standing) waves: vibrations of air columns determining wavelength and speed of sound resonance tube resonance and sonometer

As an empty vessel is filled with water, its fundamental frequency

  1. Increases

  2. Decreases

  3. Remains the same

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

   An empty vessel with a base is like a closed organ pipe , and fundamental frequency of a closed organ pipe is given by ,

       $n _{1}=v/4l$ ,
where $l=$ length of air column in pipe (height of pipe) ,
  from above we get,    
       $n _{1}\propto 1/l$ ,
when the empty vessel is filled with water , the length of air column $l$ in pipe decreases and therefore fundamental frequency $n _{1}$ increases as length of air column and fundamental frequency are inversely proportional to each other .

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

The frequency of a whistle is 200 Hz. It is approaching to stationary observer with a speed 1/3 the speed of sound. The frequency of sound as heard by the observer will be 

  1. $450 Hz$

  2. $300 Hz$

  3. $400 Hz$

  4. $425 Hz$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the Doppler effect formula for a source approaching a stationary observer: f' = f * (v / (v - vs)). Given vs = v/3, f' = 200 * (v / (v - v/3)) = 200 * (v / (2v/3)) = 200 * (3/2) = 300 Hz.

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

The amplitude of vibration of the particles of air through which a sound wave of intensity $2.0 \times 10 ^ { - 6 } \mathrm { Wm } ^ { - 2 }$ and frequency $1.0 kHz$ is passing - (Density of air = 1.2 $k g m ^ { - 3 }$  and speed of sound in air = 330 $m s ^ { - 1 }$ is)

  1. $4.4 \times 10 ^ { - 8 } m$

  2. $1.6 \times 10 ^ { - 8 } m$

  3. $2.4 \times 10 ^ { - 6 } m$

  4. $1.8 \times 10 ^ { - 6 } m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The intensity of a sound wave is given by I = 2 * pi^2 * f^2 * A^2 * rho * v. Rearranging for amplitude A: A = sqrt(I / (2 * pi^2 * f^2 * rho * v)). Plugging in values: I = 2e-6, f = 1000, rho = 1.2, v = 330. A = sqrt(2e-6 / (2 * 9.87 * 1e6 * 1.2 * 330)) = 1.6e-8 m.

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

The longitudinal waves travel in a coiled spring at a rate of 10 m/s. The distance between two consecutive compressions is 25cm. What is the frequency of the waves?

  1. 25Hz

  2. 10Hz

  3. 40Hz

  4. 250Hz
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Answer is C.

A sound wave has a speed that is mathematically related to the frequency and the wavelength of the wave. The mathematical relationship between speed, frequency and wavelength is given by the following equation.
Speed = Wavelength * Frequency. That is, Frequency = Speed / Wavelength.
In this case, the frequency is 140 per second and wavelength is 25 cm, that is, 0.25 m.
Therefore, Frequency = 10 / 0.25  = 40 Hz.
The frequency of the wave is 40 Hz.

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is $4.2$ $MH _z$. The speed of sound  in a tissue is $1.7$ ${km/s}$. The wavelength of sound in tissue is close to

  1. $4\times 10^{-4}$ $m$

  2. $8\times 10^{-4}$ $m$

  3. $4\times 10^{-3}$ $m$

  4. $8\times 10^{-3}$ $m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given:
Frequency $(f)=4.2$ $MH _z = 4.2\times 10^{6}$ $H _z$
Speed in tissue $(v)=1.7$ ${km/s} = 1700$ ${m/s}$
$\therefore$ Wavelength $=\lambda \times f=v$
$\lambda=\cfrac{v}{f}=\cfrac{1700}{4.2\times 10^{6}}=4\times 10^{-4}$ $m$

Multiple choice physics superposition of waves-2: stationary (standing) waves: vibrations of air columns determining wavelength and speed of sound resonance tube resonance and sonometer

A resonance tube apparatus is employed to.

  1. Investigate the dependence of velocity of sound in air upon temperature

  2. Verify the laws of vibrating strings

  3. Study beats

  4. Determine the velocity of sound in air
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A resonance tube is a classic laboratory apparatus used to determine the speed of sound in air by measuring the lengths of air columns that resonate with a tuning fork of known frequency.

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

Let ${ n } _{ 1 }$ and ${ n } _{ 2}$ be the two slightly different frequencies of two sound waves. The time interval between waxing and immediate next waning is ..........

  1. $\cfrac { 1 }{ { n } _{ 1 }-{ n } _{ 2 } } $

  2. $\cfrac { 2 }{ { n } _{ 1 }-{ n } _{ 2 } } $

  3. $\cfrac { { n } _{ 1 }-{ n } _{ 2 } }{ 2 } $

  4. $\cfrac { 1 }{ { 2(n } _{ 1 }-{ n } _{ 2 }) } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Beat frequency during constructive interference(waxing) is ($n _1-n _2$)
Beat frequency during destructive interference (waning) is ($n _1-n _2$)
The combination of two waves will give beat frequency as $2(n _1-n _2)$
Now ,the number of beats produced per one second is defined as the reciprocal of difference in frequencies two sound waves which produce waxing and waning.
$\therefore\ $ Time interval between waxing and immediate waning is $=\dfrac{1}{2(n _1-n _2)}$