Tag: physics

Questions Related to physics

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?

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

  1. 25Hz

  2. 10Hz

  3. 40Hz

  4. 250Hz

Show answer
Correct Option: C
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.

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

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

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

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

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

  4. $8\times 10^{-3}$ $m$

Show answer
Correct Option: A
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 $=\lambda \times f=v$
$\lambda=\cfrac{v}{f}=\cfrac{1700}{4.2\times 10^{6}}=4\times 10^{-4}$ $m$

A resonance tube apparatus is employed to.

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

  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

Show answer
Correct Option: D

The audible range of frequency of sound waves for human beings is _____.

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

  1. 10 Hz to 10,000 Hz

  2. 20 Hz to 20,000 Hz

  3. 5 Hz to 50,000 Hz

  4. 50 Hz to 20,000 Hz

Show answer
Correct Option: B

If the frequency of human heart is $1.25$ Hz, the number of heart beats in $1$ minute is

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

  1. $65$

  2. $75$

  3. $80$

  4. $90$

Show answer
Correct Option: B
Explanation:

Beat frequency of heart = $1.25$Hz.
$\therefore$ Number of beats in $1$ minute = $1.25 \times 60 = 75$.

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 ..........

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

  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 }) } $

Show answer
Correct Option: D
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 $=\dfrac{1}{2(n _1-n _2)}$ 

In a resonating air column, the first booming sound is heard when the length of air column is $10\ cm$. The second booming sound will be heard when length is:

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

  1. $20\ cm$

  2. $30\ cm$

  3. $40\ cm$

  4. None of the above

Show answer
Correct Option: B
Explanation:

Booming sound indicates that at that length, $l _1$, the air column is in resonance with the given frequency.
and that length is,
$l _1= \lambda /4=10$
or, $\lambda = 40cm$
The next resonance length will be :
$l _2=3\lambda/4=30 cm$

In Kundt's tube, when waves of frequency $10^3\space Hz$ are produces the distance between five consecutive nodes is $82.5\space cm$. The speed of sound in gas filled in the tube will be

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

  1. $660\space ms^{-1}$

  2. $330\space ms^{-1}$

  3. $230\space ms^{-1}$

  4. $100\space ms^{-1}$

Show answer
Correct Option: B
Explanation:

$\quad \displaystyle\frac{5\lambda}{2} = 82.5\space cm$

$\quad \lambda = 33\space cm\quad and \quad v = f\lambda = 330\space ms^{-1}$ 

The frequency of a fork is $500$Hz. Velocity of sound in air is $350$ $ms^{-1}$. The distance through which sound travel by the time the fork makes $125$ vibrations is?

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

  1. $87.5$m

  2. $700$m

  3. $1400$m

  4. $1.75$m

Show answer
Correct Option: A
Explanation:

$wavelength=\dfrac { velocity }{ frequency } $ 

$=\dfrac { 350 }{ 500 } =\dfrac { 7 }{ 10 } $
Distance traveled in $125$ vibrations
$=$wavelength$\times$ no of vibrations
$=\dfrac { 7 }{ 10 } \times 125$
 $=87.15m$

Frequency of tuning fork $A$ is $256\ Hz.$ It produces four beats/sec with tuning fork $B.$ When wax is applied at tuning fork $B$ then $6$ beats/sec are heard. By reducing little amount of wax $4$ beats/sec are heard. Frequency of $B$ is : 

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

  1. $250\ Hz$

  2. $252\ Hz$

  3. $260\ Hz$

  4. $256\ Hz$

Show answer
Correct Option: B
Explanation:

Let the unknown frequency of the tuning fork be x.

So, according to the given data when no waxed, its frequency must be,

$x=256\pm 4$  to produced a beat of $4\ beats /sec$.

We know, the frequency of a tuning fork decreases as it is waxed.

So, to produce $6\  beats/s$, after being waxed, the frequency of the tuning fork must be

  $ x=256-4 $

 $ x=252\,Hz $

Hence, the frequency of $B$ is $252\ Hz$