Tag: beats in sound waves

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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.

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

$y _1 = A cos (2f _1t)$     and $y _2 = A cos (2 f _2t),$, then $y _{total} $ is

  1. $y _{total} = y _1 + y _2 = A {cos (2 f _1t) + cos (2 f _2t)}$
  2. $y _{total} = y _1 - y _2 = A {cos (2 f _1t) - cos (2 f _2t)}$
  3. $y _{total} =\dfrac{ y _1}{ y _2} = A\dfrac{{cos (2 f _1t)}}{{cos (2 f _2t)}}$
  4. $y _{total} = y _1 \times y _2 = A {cos (2 f _1t) \times cos (2 f _2t)}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Resultant  $(y _{total})$ of the two waves is equal to the superposition of the waves and is given by,
$y _{total}  = y _1+y _2$
$\therefore$  $y _{total} = A \ cos(2f _1t)+ A \ cos(2f _2t)$

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 of wavelengths 99 cm and 100 cm both travelling with velocity 396 m/s are made of interfere. The number of beats produced by them per second are

  1. $1$
  2. $2$
  3. $4$
  4. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} Velocity\, \, of\, \, wave\, \, V=n\lambda  \ Where\, \, n=frequency\, \, of\, \, wave\, \,  \ \Rightarrow n=\frac { v }{ \lambda  }  \ { n _{ 2 } }=\frac { { { v _{ 2 } } } }{ { { \lambda _{ 2 } } } } =\frac { { 396 } }{ { 100\times { { 10 }^{ -2 } } } } =396Hz \ no.\, \, of\, \, beats\, \, ={ n _{ 1 } }-n _2\, =4 \end{array}$