Tag: stationary waves

Quizzes (7) Questions (39)

Questions Related to stationary waves

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 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 $=\dfrac{1}{2(n _1-n _2)}$ 
Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

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

  1. $660\space ms^{-1}$
  2. $330\space ms^{-1}$
  3. $230\space ms^{-1}$
  4. $100\space ms^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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?

  1. $87.5$m
  2. $700$m
  3. $1400$m
  4. $1.75$m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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 : 

  1. $250\ Hz$
  2. $252\ Hz$
  3. $260\ Hz$
  4. $256\ Hz$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $252\ Hz$

 

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

Two pendulums of length $1.21m$ and $1.0m$ start vibrating. At some instant, the two are in the mean position in same phase. After how many vibrations of the longer pendulum, the two will be in phase?

  1. $10$
  2. $11$
  3. $20$
  4. $21$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time period of a pendulum is T = 2*pi*sqrt(L/g). T1/T2 = sqrt(L1/L2) = sqrt(1.21/1.0) = 1.1 = 11/10. For them to be in phase, n1*T1 = n2*T2. n1/n2 = T2/T1 = 10/11. After 11 vibrations of the longer pendulum (T1), the shorter one (T2) will have completed 10 vibrations.