Tag: waves

Questions Related to waves

If n$ _{1},n _{2},n _{3}$ are the three  fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency '$n$' of the string is given by 

  1. $\sqrt{n}=\sqrt{n _{1}}+\sqrt{n _{2}}+\sqrt{n _{3}}$

  2. $\displaystyle \dfrac{1}{\sqrt{n}}=\dfrac{1}{\sqrt{n _{1}}}+\dfrac{1}{\sqrt{n _{2}}}+\dfrac{1}{\sqrt{n _{3}}}$

  3. $n=n _{1}+n _{2}+n _{3}$

  4. ${\dfrac{1}{n}}=\displaystyle \dfrac{1}{n _{1}}+\dfrac{1}{n _{2}}+\dfrac{1}{n _{3}}$


Correct Option: D
Explanation:

Total length of string is $l=l _{1}+l _{2}+l _{3}$
but $f\propto \dfrac{1}{l}$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{f _{1}}+\dfrac{1}{f _{2}}+\dfrac{1}{f _{3}}$

To increase the frequency by $20\%$, the tension in the string vibrating on a Sonometer has to be increased by

  1. $44\%$

  2. $33\%$

  3. $22\%$

  4. $11\%$


Correct Option: A
Explanation:

frequency increased by $20\%
$$\Rightarrow f^{'}=\dfrac{6}{5}f$
$\therefore \sqrt{T^{'}}=\dfrac{6}{5}\sqrt{T}$
$\sqrt{T^{'}}=\sqrt{\dfrac{144}{100}}T$
$\therefore$ Tension is to be increased by $44\%$

An iron load of $2 kg$ is suspended in air from the free end of a sonometer wire of length one meter. A tuning fork of frequency $256 Hz$ is in resonance with $1/\sqrt{7}$ times the length of the sonometer wire. If the load is immersed in water, the length of the wire in meter that will be in resonance with the same tuning fork is :


(Specific gravity of iron $= 8$)

  1. $\sqrt{8}$

  2. $\sqrt{6}$

  3. $\displaystyle \dfrac{1}{\sqrt{6}}$

  4. $\displaystyle \dfrac{1}{\sqrt{8}}$


Correct Option: D
Explanation:

$f\propto \dfrac{1}{l}\sqrt{\dfrac{T}{\mu}}$
$\Rightarrow l\propto \sqrt{T}$
$\therefore \dfrac{l _{1}}{l _{2}}=\sqrt{\dfrac{T _{1}}{T _{2}}}=\sqrt{\dfrac{8}{7}}$
$\dfrac{1}{\sqrt{7}l _{2}}=\dfrac{\sqrt{8}}{\sqrt{7}}$
$\Rightarrow l _{2}=\dfrac{1}{\sqrt{8}}$

The maximum particle velocity is $8$ times the wave velocity of a progressive wave. If the amplitude of the particle is $"a"$. The phase difference between the two particles seperated by a distance of $""x"$ is 

  1. $\frac{x}{a}$

  2. $\frac{{8x}}{a}$

  3. $\frac{{3a}}{x}$

  4. $\frac{{3\pi x}}{a}$


Correct Option: B

The particle of a medium vibrates about their mean position whenever a wave travels through that medium. The phase difference between the vibrations of two such particles

  1. varies with time only

  2. varies with distance separating them only

  3. varies with time as well as distance

  4. is always zero


Correct Option: B
Explanation:

The phase difference between the vibrations of two particles of the medium is given by :

            $\Delta \phi=\dfrac{2\pi}{\lambda}\Delta x$ 
it is clear that phase difference varies as the path difference between the particles varies, which is the distance, separating the particles.

The phase difference between two points separated 0.8 m in a wave of frequency 120 HZ is 0.5 $\pi $ the value velocity is

  1. 144 ${ ms }^{ -1 }$

  2. 384 ${ ms }^{ -1 }$

  3. 256 ${ ms }^{ -1 }$

  4. 720 ${ ms }^{ -1 }$


Correct Option: B

Two Waves of amplitudes ${ A } _{ 0 }$ and $x{ A } _{ 0 } $ pass through a region. If x >1, the difference in the maximum and minimum resultant amplitude possible is

  1. $(x+1){ A } _{ 0 }$

  2. $(x-1){ A } _{ 0 }$

  3. $2x{ A } _{ 0 }$

  4. $2{ A } _{ 0 }$


Correct Option: D

The equation of a wave is given by $Y\, =\, 5\, sin\, 10 \pi\, (t\, -\, 0.01x)$ along the x-axis. (All the quantities are expressed in SI units}. The phase difference the points separated by a distance of 10 m along x-axis is

  1. $\displaystyle \frac{\pi}{2}$

  2. $\pi$

  3. $2 \pi$

  4. $\displaystyle \frac{\pi}{4}$


Correct Option: B
Explanation:

For the given wave, $k=0.1\pi=\dfrac{2\pi}{\lambda}$
$\implies \lambda=20m$

Thus phase difference between two points separated by 10m is $\dfrac{2\pi}{\lambda}(x _2-x _1)$
$=\dfrac{2\pi}{\lambda}\times 10m=\pi$
Hence correct answer is option B.

A uniform rope of length $L$ and mass ${m _1}$ hangs vertically from a rigid support . A block of mass ${m _2}$ is attached to the free end of the rope. A transverse pulse of wavelength ${\lambda _1}$ is produced at  the lower end of the rope . the wavelength of the pulse when it reaches the top of the rope is ${\lambda _2}$. The ratio $\frac{{{\lambda _1}}}{{{\lambda _2}}}$ is:

  1. $\sqrt {\dfrac{{{m _1}}}{{{m _2}}}} $

  2. $\sqrt {\dfrac{{{m _1} + {m _2}}}{{{m _2}}}} $

  3. $\sqrt {\dfrac{{{m _2}}}{{{m _1}}}} $

  4. $\sqrt {\dfrac{{{m _1} + {m _2}}}{{{m _1}}}} $


Correct Option: D

When a wave travels in a medium, the particle displacement is given by y(xt)=0.03 sin $\pi $ (2t-0.01 x) where y and x are meters and t in seconds. The phase difference, at a given instant of time between two particle 25 m. apart in the medium, is 

  1. $\frac{\pi }{8}$

  2. $\frac{\pi }{4}$

  3. $\frac{\pi }{2}$

  4. $\pi$


Correct Option: A