Tag: superposition of waves-1: interference and beats

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Questions Related to superposition of waves-1: interference and beats

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Two waves of intensities $I$ and $4I$ superimpose. The minimum and maximum intensities will respectively be

  1. $I,\space 9I$
  2. $3I,\space 5I$
  3. $I,\space 5I$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The Intensity of the wave is directly proportional to the square of its amplitude.
$I \propto A^{2}$;
$I = cA$$^2$;   '$c$' is an arbitrary constant.
So, if a wave with Intensity $I$ has an amplitude of $A(A{ _{1}}$)
A wave with an Intensity of $4I$ would respectively have an amplitude of $2A$($A{ _{2}}$)

If 2 waves with amplitudes $A{ _{1}}$,$A{ _{2}}$ are superimposed, the resultants would be
Maximum of $A{ _{1}}$ + $A{ _{2}} = 3A$ (Constructive Interference)
Minimum of $|A{ _{1}}$ - $A{ _{2}} |   = A $ (Destructive Interference)
The wave of amplitude $3A$ would have an Intensity of $9I$
The wave of amplitude $A$ would have an Intensity of $I$
Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

For a wave displacement amplitude is $10^{-8} m$ density of air $1.3 kg m^{-3}$ velocity in air $340 ms^{-1}$ and frequency is 2000 Hz.The average intensity of wave is

  1. $5.3\times 10^{-4} Wm^{-2}$
  2. $5.3\times 10^{-6} Wm^{-2}$
  3. $3.5\times 10^{-8} Wm^{-2}$
  4. $3.5\times 10^{-6} Wm^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Intensity I = 2 * pi^2 * f^2 * A^2 * rho * v. Plugging in values: I = 2 * (3.14)^2 * (2000)^2 * (10^-8)^2 * 1.3 * 340. I = 2 * 9.86 * 4 * 10^6 * 10^-16 * 1.3 * 340 = 5.3 * 10^-4 W/m^2.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Two sound waves of equal intensity $I$ superimpose at point $P$ in $90^{\small\circ}$ out of phase. The resultant intensity at point $P$ will be

  1. $4I$
  2. $\sqrt2I$
  3. $2I$
  4. $I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Amplitude of the resultant wave is:
$A _R=\sqrt{A^2 _1+A^2 _2+2A _1A _2\cos(\theta)}$.

Here, $A _1=A _2=A \text{ and } \theta = \pi/2$

So, $A _R=A\sqrt{2(1+\cos(\theta))}=2A\cos(\theta/2)=2A\cos(\pi/4)=\sqrt{2}A$ 

$\Rightarrow I _R=|A _R|^2=2A^2=2I$

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

A wave of frequency 500$\mathrm { Hz }$ travels between $\mathrm { x }$and $\mathrm { Y }$ and travel a distance of 600$\mathrm { m }$ in 2$\mathrm { sec }$ . between $X$ and $Y .$ How many wavelength are therein distance $X Y$ :

  1. 1000

  2. 300

  3. 180

  4. 2000
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Velocity v = distance/time = 600m / 2s = 300 m/s. Wavelength lambda = v/f = 300/500 = 0.6 m. Number of wavelengths = distance / lambda = 600 / 0.6 = 1000.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Four independent waves are represented by the equations :
$y _1 = a _1\  sin\  \omega t$
$y _2 = a _2\ sin\  \omega t$
$y _3 = a _3\ cos\  \omega t$
$y _4 = a _4\ sin\  (\omega t + \pi/3)$ 
Then the waves for which phenomenon of interference will be observed are - 

  1. 1 and 3

  2. 1 and 4

  3. all 1, 2, 3 and 4

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

Interference is observed between waves of the same frequency. Waves 1 (sin wt) and 3 (cos wt = sin(wt + pi/2)) have the same frequency (omega). Wave 2 also has frequency omega, but 1 and 3 are the standard pair for demonstrating interference.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Two sinusoidal plane waves same frequency having intensities $I _0 $ and $ 4I _0 $ are travelling in same direction. The resultant intensity at a point at which waves meet with a phase difference of zero radian is

  1. $ I _0$
  2. $5 I _0$
  3. $9 I _0$
  4. $3 I _0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let, $I _1=I _0  $ and $  I _2=4I _0 $

Resultant intensity, $I=I _1+I _2+2\sqrt{I _1I _2} cos\phi $
                                   $= I _0+4I _0+2\sqrt{I _04I _0} cos0^{\circ} \ =9I _0 $

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

If the ratio of maximum to minimum intensity in beat is 49, then the ratio of amplitudes of two progressive wave trains

  1. 7:1

  2. 4:3

  3. 49:1

  4. 16:9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\dfrac{I _{max}}{I _{min}}=\dfrac{(\sqrt{I _1}+\sqrt{I _2})^2}{(\sqrt{I _1}-\sqrt{I _2})^2}=49$

$\dfrac{(\sqrt{I _1}+\sqrt{I _2})}{(\sqrt{I _1}-\sqrt{I _2})}=7$

$\sqrt{I _1}+\sqrt{I _2}=7(\sqrt{I _1}-\sqrt{I _2})$

$\dfrac{\sqrt{I _1}}{\sqrt{I _2}}=\dfrac{4}{3}$

$\dfrac{a}{b}=\dfrac{4}{3}$

Here, $a =\sqrt{I _1}$ and $b =\sqrt{I _2}$, where a and b are the amplitudes of the two progressive waves.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

If the intensities of two interfering waves be $ I _1 $ and $ I _2  $, the contrast between maximum and minimum intensity is maximum, when

  1. $I _1 > > I _2$
  2. $I _1 < < I _2$
  3. $I _1 = I _2$
  4. either $I _1$ or $I _2$ is zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$I _{max}=(\sqrt{I _1}+\sqrt{I _2})^2$
$I _{min}=(\sqrt{I _1}-\sqrt{I _2})^2$
Contrast is maximum, when $I _{min}=0$ ie. $I _1=I _2$

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

If the phase difference between two sound waves of wavelength $  \lambda  $ is $  60^{\circ} $, the corresponding path difference is

  1. $ \frac{\lambda}{6} $
  2. $ \frac{\lambda}{2} $
  3. $ \lambda 2 $
  4. $ \frac{\lambda}{4} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relationship between phase difference (delta phi) and path difference (delta x) is delta phi = (2 * pi / lambda) * delta x. Given delta phi = 60 degrees = pi/3 radians, pi/3 = (2 * pi / lambda) * delta x. Solving for delta x gives delta x = lambda / 6.