Tag: distinction between interference and beats

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Questions Related to distinction between interference and beats

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

Two sound sources are moving in opposite direction with velocity $v _1$ and $v _2$ $(v _1>v _2)$. Both are moving away from a stationary observer.the frequency of both the source is $900\ Hz$. What is the value of $v _1 - v _2 $  so that the beat frequency observed will be $6\ Hz$  ?


Speed of sound =$300\ ms^{-1}$

  1. $1\ ms^{-1}$
  2. $4\ ms^{-1}$
  3. $3\ ms^{-1}$
  4. $2\ ms^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$f _1 = 900(\dfrac{300}{300+v _1})$

$\implies f _1= 900(1+ \dfrac{v _1}{300})^{-1}$
$\implies f _1 = 900 - 3v _1$
Similarly 
$f _2 = 900 - 3v _2$
So,
$f _1 -f _2 = 6$
$3(v _1 - v _2)= 6$
$\implies v _1 - v _2 = 2\ ms^{-1}$

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

A sources of sonic oscillations with frequency n= $1700$ Hz and a receiver are located on the same normal to a wall. Both the source and receiver are stationary, and the wall recedes from the source with velocity u= $6.0$ cm/s. Find the beat frequency registred by the receiver. The velocity of sound is equal to $v= 340$ m/s.

  1. $0.2$ Hz
  2. $0.3$ Hz
  3. $0.4$ Hz
  4. $0.6$ Hz
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The wall acts as a moving reflector. The frequency of the sound reflected by the wall is f' = f * (v + u) / (v - u). The beat frequency is the difference between the reflected frequency and the source frequency, which simplifies to f_beat = f * (2u / (v - u)). Plugging in f=1700, u=0.06 m/s, and v=340 m/s gives 1700 * (0.12 / 339.94), which is approximately 0.6 Hz.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

Two sound sources (of same frequency ) are placed at distance of 100 meter. An observer, when moving between both sources, hears 44 beats per second. The distance between sound source is now changed to 400 meter then the beats/second heard by observer will be  :

  1. 2

  2. 4

  3. 8

  4. 16
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

Two open organ pipes 80 and 81 cm long found to give 26 beats in 10 sec, when each is sounding its fundamental note. Then the velocity of sound in air is

  1. 337 $m s ^ { - 1 }$
  2. 370 $m s ^ { - 1 }$
  3. 345 $m s ^ { - 1 }$
  4. 350 $m s ^ { - 1 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The beat frequency is 26 beats / 10 sec = 2.6 Hz. For an open pipe, f = v / (2L). The difference in frequencies is f1 - f2 = (v/2) * (1/L1 - 1/L2) = 2.6. Substituting L1 = 0.80 m and L2 = 0.81 m, we get (v/2) * (0.01 / (0.80 * 0.81)) = 2.6, which solves to v = 336.96 m/s, approximately 337 m/s.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

Two monochromatic light waves of amplitudes $A$ and $2A$ interfering at a point, have a phase difference of ${60^0}.$ The intensity at that point will be  proportional to :

  1. $3{A^2}$
  2. $5{A^2}$
  3. $7{A^2}$
  4. $9{A^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The resultant intensity I is given by I = I1 + I2 + 2 * sqrt(I1 * I2) * cos(phi). Since intensity is proportional to amplitude squared, I1 = k * A^2 and I2 = k * (2A)^2 = 4 * k * A^2. With phi = 60 degrees, cos(60) = 0.5. Thus, I = k * A^2 + 4 * k * A^2 + 2 * sqrt(k * A^2 * 4 * k * A^2) * 0.5 = 5 * k * A^2 + 2 * (2 * k * A^2) * 0.5 = 7 * k * A^2.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

Two sound waves with wavelength $5$m and $5.5$ m respectively. each propoggate in a gas with velocity $300$ m/s. we expect the following number of beats per second

  1. 12

  2. 0

  3. 1

  4. 6
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Frequency f = v / lambda. f1 = 300 / 5 = 60 Hz. f2 = 300 / 5.5 = 54.54 Hz. The beat frequency is |f1 - f2| = |60 - 54.54| = 5.46 Hz, which is approximately 6 Hz.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

Two waves are approaching each other with a velocity of $16\, m/s$ and frequency $n$. the distance between two consecutive nodes is 

  1. $\dfrac{16}{n}$
  2. $\dfrac{8}{n}$
  3. $\dfrac{n}{16}$
  4. $\dfrac{n}{8}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The wavelength lambda is given by velocity divided by frequency, so lambda = 16 / n. The distance between two consecutive nodes in a standing wave is equal to half of the wavelength, which is lambda / 2 = 8 / n.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

A body is walking away from a wall towards an observe at a speed of 1 m/s and blows a whistle whose frequency is 680 Hz. The number of beats heard by the observe per second is approximately.(velocity of sound in air = 340 m/s)

  1. 4

  2. 8

  3. 2

  4. zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sound reflects off the wall. The wall acts as a source moving toward the observer at 1 m/s. The frequency heard is f' = f * (v + u) / (v - u). With f = 680, v = 340, u = 1, f' = 680 * (341 / 339) = 684 Hz. The beat frequency is f' - f = 684 - 680 = 4 Hz.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

Two sound waves of equal intensity I produce beats . The maximum intensity of sound produced in beats will be

  1. I

  2. 4I

  3. 2I

  4. I/2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The maximum intensity of sound in interference or beats is given by I_max = (sqrt(I1) + sqrt(I2))^2. Since both waves have equal intensity I, I_max = (sqrt(I) + sqrt(I))^2 = (2*sqrt(I))^2 = 4I.