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

Suppose displacement produced at some point $P$ by a wave is $y _1=a cos \omega t$ and by another wave is $y _2=a cos \omega t$.Let $I _0$ represents intensity produced by each one of individual wave, then resultant intensity due to overlapping of both wave is

  1. $I _0$
  2. $2I _0$
  3. $\dfrac{I _0}{2}$
  4. $4I _0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$y _{res}$ at point $P=y _1+y _2\=a\cos \omega t+a\cos\omega t\=2\cos\omega t$

Now amplitude $=2a$
Since  intensity $\propto$(amplitude)$^2$
So, $\cfrac{I _{new}}{I _0}=\cfrac{4a^2}{a^2}\\implies I _{new}=4I _0$

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

Two waves $Y _{1}=a\sin \omega t$ and $Y _{2}=a\sin (\omega t+\delta)$ are producing interference, then resultant intensity is-

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

The resultant intensity of two waves with equal amplitude 'a' and phase difference 'delta' is I = I1 + I2 + 2*sqrt(I1*I2)*cos(delta). With I1 = I2 = k*a^2, I = 2*k*a^2(1 + cos(delta)) = 4*k*a^2*cos^2(delta/2). Assuming k=1, the result is 4*a^2*cos^2(delta/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

Sounds from two identical $S _1$ and $S _2$ reach a point  P. When the sounds reach directly, and in the same phase, the intensity at $P$ is $I _0$. The power of $S _1$ is now reduced by $64\%$ and the phase difference between $S _1$ and $S _2$ is varied continuously. The maximum and minimum  intensities recorded at P are mow $I _{max}$ and $I _{min}$ 

  1. $I _{max}=0.64I _0$
  2. $I _{min}=0.36I _0$
  3. $\dfrac{I _{max}}{I _{min}}=16$
  4. $\dfrac{I _{max}}{I _{min}}=\dfrac{1.64}{0.36}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial intensity I0 = (sqrt(I1) + sqrt(I2))^2. Since S1 and S2 are identical, I1 = I2 = I, so I0 = (2*sqrt(I))^2 = 4I. Power of S1 reduced by 64 percent means new I1 = 0.36I. New Imax = (sqrt(0.36I) + sqrt(I))^2 = (0.6 + 1)^2 * I = 2.56I. Since I0 = 4I, I = I0/4. Thus Imax = 2.56 * (I0/4) = 0.64I0.

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

Path difference between two waves from a coherent sources is 5 nm at a  point P. Wavelength of these waves is 100 $\mathring { A } $. Resultant intensity at point P if intensity of sources is $l _0$ and $4l _0$

  1. Zero
  2. $l _0$
  3. $5l _0$
  4. $3l _0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} \Delta x=5\times { 10^{ -9 } }m \ \Delta \phi =\frac { { 2\pi  } }{ \lambda  } \Delta x \ =\frac { { 2\pi \times 5\times { { 10 }^{ -9 } } } }{ { 100\times { { 10 }^{ -10 } } } }  \ =\pi  \ Now, \ I={ I _{ 0 } }+4{ I _{ 0 } }+2\sqrt { { I _{ 0 } } } \sqrt { 4{ I _{ 0 } } } \cos  \phi  \ =5{ I _{ 0 } }+4{ I _{ 0 } }\left[ { \cos  \pi  } \right]  \ ={ I _{ 0 } } \ \therefore resula\tan  t\, \, \, { { intensity } }\, ={ I _{ 0 } } \ Hence,\, option\, \, B\, \, is\, the\, correct\, answer. \end{array}$

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 laser beam can be focussed on an area equal to the square of its wavelength. A He-Ne laser radiates energy at the rate of $1\,nW$ and its wavelength is $632.8\,nm$.The intensity of foucussed beam will be   

  1. $1.5 \times {10^{13}}\,W/{m^2}$
  2. $0.25 \times {10^{4}}\,W/{m^2}$
  3. $3.5 \times {10^{17}}\,W/{m^2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A laser beam can be focused on an area equal to the square of its wavelength.

A He-Ne laser radial co energy 
at the rate$=1nW$
wavelength$=6.32.8nm$
The intensity of focused beam willbe
Area through which energy of beam passes
$=(6.328\times10^{-7})=4\times10^{-13}m^2\I=\cfrac{P}{A}=\cfrac{10^{-9}}{4\times10^{-13}}\ \quad=0.25\times10^4W/m^2$

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 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.