Tag: waves

Quizzes (32) Questions (233)

Questions Related to waves

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

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. ${\dfrac{1}{n}}=\displaystyle \dfrac{1}{n _{1}}+\dfrac{1}{n _{2}}+\dfrac{1}{n _{3}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

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. $11\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $44\%$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

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. $\displaystyle \dfrac{1}{\sqrt{8}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

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. $2{ A } _{ 0 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Maximum amplitude A_max = A1 + A2 = (x + 1) * A_0. Minimum amplitude A_min = |A1 - A2| = (x - 1) * A_0. The difference is A_max - A_min = (x + 1) * A_0 - (x - 1) * A_0 = 2 * A_0.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

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. $\displaystyle \frac{\pi}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

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}}}} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

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. $\pi$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The phase difference delta phi is given by (2pi / lambda) * delta x, or from the wave equation k * delta x. Here k = 0.01 pi, so delta phi = 0.01 pi * 25 = 0.25 pi = pi / 4 radians.