Tag: reflection of waves

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Multiple choice physics wave motion reflection of waves

A wave travels on a light string. The equation of the waves is $Y\, = \,A\, sin\,(kx\,-\,\omega\,t+\,30^{\circ})$. It is reflected from a heavy string tied to end of the light string at x = 0 . If 64% of the incident energy is reflected then the equation of the reflected wave is  

  1. $Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ}\,+\,180^{\circ})$
  2. $Y\, =\,0.8 \,A\, sin\,(kx\,+\,\omega\,t\,+\,30^{\circ}\,+\,180^{\circ})$
  3. $Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,-\,30^{\circ})$
  4. $Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ})$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

There are three things we need to take into account:

  • Energy transfer
  • Change in velocity
  • Change in phase
We know that power delivered is proportional to $A^{2}$
Hence if power(energy) reduces to 64%. We get that Amplitude must reduce to 80% or 0.8A.
Now the reflected wave is moving in the opposite direction. (velocity is negative now).
Also because of the hard soft boundary reflection (there is a phase lag of $180^{\circ}$
Hence the new equation becomes:
$y = 0.8A sin(kx + \omega t + 30^{\circ} + 180^{\circ})$
Hence option B.

Multiple choice physics wave motion reflection of waves

A pulse of a wave train travels along a stretched string and reaches the fixed end of the string. It will be reflected back with :

  1. a phase change of ${180}^{o}$ with velocity reversed

  2. the same phase as the incident pulse with no reversal of velocity
  3. a phase change of ${180}^{o}$ with no reversal of velocity

  4. the same phase as the incident pulse but with velocity reversed
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A pulse of wave train when travels along a stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium and the reflected ray suffers a phase change of $\pi$ with the incident wave but wave velocity after reflection reverses its direction.

Multiple choice physics wave motion reflection of waves

A wave of length $2m$ is superposed on its reflected wave to form a stationary wave. A node is located at  $ x=3m$ The next node will be located at  $x=$

  1. $4m$
  2. $3.75m$
  3. $3.50m$
  4. $3.25m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since wave length is $2m$, half of wavelength is $1m$. Node forms after each half of wave length. So, node will be formed at each $1 m$
So, as node is formed at $3 m$
next node will be formed at $3+1=4\ m.$

Multiple choice physics wave motion reflection of waves

A sound wave of frequency $1360 Hz$ falls normally on a perfectly reflecting  wall.  The shortest distance from the wall at which the air particles have maximum amplitude of vibration is ($v = 340 m/s$)

  1. $25 cm$
  2. $6.25 cm$
  3. $62.5 cm$
  4. $2.5 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$v=f\lambda$
$\lambda=\dfrac{340}{1360}=\dfrac{1}{4}=0.25m$
$=25cm$
As the end is a node, shortest distance at which antinode is formed is $\dfrac{\lambda}{4}=6.25cm$

Multiple choice physics wave motion reflection of waves

A string fixed at one end only is vibrating in its third harmonic. The wave function is $y(x,t) = 0.02 sin(3.13x) cos(512t)$, where y and x are in metres and t is in seconds. The nodes are formed at positions

  1. (0 m, 2 m)

  2. (0.5 m, 1.5 m)

  3. (0 m, 1.5 m)

  4. (0.5 m, 2 m)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Node in formed only at the finest end of the string and the free end acts as an anti node.

In the third harmonics two nodes are formed:

$y\left( x,t \right) =0.02\sin { \left( 3.13x \right) \cos { \left( 512t \right)  }  } $

Standard equation given by

$y\left( x,t \right) =2a\sin { \left( \cfrac { 2\pi  }{ \lambda  } x \right)  } \cos { \left( 2\pi vt \right)  } $

Comparing both equation, we get

$3.13x\quad =\cfrac { 2\pi  }{ \lambda  } x\\ or,\quad \lambda =\cfrac { 2\lambda  }{ 3.13 } \\ \quad \quad \quad \quad =2 m (approx)$

The nodes are formed at $\cfrac { \lambda  }{ 4 } =0.5$ from origin and at $\cfrac { 3\lambda  }{ 4 } =1.5$ from origin.

 

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

A string is under tension so that its length is increased by $1/n$ times its original length. The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be

  1. $1:n$
  2. ${n}^{2}:1$
  3. $\sqrt{n}:1$
  4. $n:n+1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The frequency of longitudinal vibrations depends on the speed of sound in the material, while transverse vibrations depend on tension. Calculating the ratio based on the extension 1/n leads to the n/(n+1) relationship.

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

Motion that moves to and fro in regular time intervals is called _________________ motion.

  1. Vibratory

  2. Translatory

  3. Rotatory

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

Sounds are made by vibrations. Some vibrations are easy to see. The vibrations that create sound must travel through a medium, such as air or water, or anything made of molecules. With each forward motion, air molecules pulse outward, pushing other air molecules and crowding them together. With each backward motion, the molecules get less crowded. The forward and backward vibration of the glass creates a chain reaction of crowded and not-so-crowded molecules that ripples through the air. This traveling vibration is called a sound wave. 
Motion that moves to and fro in regular time intervals is called vibratory or oscillatory motion.

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

When we hear a sound, we can identify its source from : 

  1. Amplitude of sound

  2. Intensity of sound

  3. Wavelength of sound

  4. Overtones present in the sound
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Answer is D.

When we hear a sound, we can identify its source from overtones present in the sound.
The fundamental is the frequency at which the entire wave vibrates. Overtones are other sinusoidal components present at frequencies above the fundamental. All of the frequency components that make up the total waveform, including the fundamental and the overtones, are called partials. Together they form the harmonic series.
Overtones which are perfect integer multiples of the fundamental are called harmonics. When an overtone is near to being harmonic, but not exact, it is sometimes called a harmonic partial, although they are often referred to simply as harmonics. Sometimes overtones are created that are not anywhere near a harmonic, and are just called partials or inharmonic overtones.

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

The length of a stretched string is $2 m$. The tension in it and its mass are $10 N$ and $0.80 kg$ respectively. Arrange the following steps in a sequence to find the third harmonic of transverse wave that can be created in the string.
(a) Find the linear mass density ($m$) using the formula, $m$ $\displaystyle = \dfrac{mass (M)  of \ the \  string}{length (l)  of \ the \  string}$
(b) Collect the data from the problem and find the length($l$) tenstion ($T$) and mass ($M$) of the stretched string.
(c) The fundamental frequency of a stretched vibrating string is given by $n$ $=\displaystyle \dfrac{1}{2l} \sqrt{\dfrac{T}{m}}$
(d) The frequency of $2^{nd}$ overtone or $3^{rd}$ harmonic is given by $n _2\displaystyle = \dfrac{3}{2l}\sqrt{\dfrac{T}{m}}=3n$.

  1. a b c d

  2. d b c a

  3. b a c d

  4. b d c a
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Collect the data from the problem and find the length ($l$), tension ($T$) and mass ($M$) of the stretched string (b). 

Find the linear mass density (m) using the formula, m $= m/l$  (a).
The fundamental frequency of a stretched vibrating string is given by, n $=\displaystyle \dfrac{1}{2l} \sqrt{\dfrac{T}{m}}$ (c).
The frequency of $2^{nd}$ overtone or $3^{rd}$ harmonic is given by $n _2=\displaystyle \dfrac{3}{2l} \sqrt{\dfrac{T}{m}}=3n$ (d)