Tag: reflection of waves

Questions Related to reflection of waves

The frequency of A note is $4$ times that of B note. The energies of two notes are equal. The amplitude of B note as compared to that of A note will be:

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

  1. double

  2. equal

  3. four times

  4. eight times

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Correct Option: A
Explanation:

$E$ is for energy, $A$ for amplitude and $f $ for frequency.

As per the problem $E _{A} = E _{B}$
Hence, $f _{A} \times A _{A}^{2} = f _{B} \times A _B^2$
 $4f _{B} \times A _{A}^{2} = f _{B} \times A _B^2$
Hence, $2A _A = A _B $

A string vibrates in 5 segment to a frequency of 480 Hz. The frequency that will cause it to vibrate in 2 segments will be

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

  1. 96 Hz

  2. 192 Hz

  3. 1200 Hz

  4. 2400 Hz

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Correct Option: B
Explanation:

5 segments implies $\lambda = \dfrac{2}{5}l$
$\nu = \dfrac{v}{\lambda} = \dfrac{5v}{2l} = 480Hz$
If the string is in 2 segments.
$\lambda = l$
$\nu = \dfrac{v}{\lambda} = \dfrac{2}{5} \dfrac{5v}{2l} = \dfrac{2}{5} 480 = 192Hz$
Hence option B is correct.

The vibrating body while playing a violin is ___________.

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

  1. wire

  2. the box of the violin

  3. both wire and box

  4. only air

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Correct Option: C

If you set up the seven overtone on a string fixed at both ends, how many nodes and antinodes are set up in it?

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

  1. $6, 5$

  2. $5, 4$

  3. $4, 3$

  4. $3, 2$

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Correct Option: C
Explanation:

Third mode of vibration or second overtone has three loops.
Its consist of $4$ nodes and $3$ antinodes.

A pipe of length $l _1$ closed at one end is kept in a chamber of gas density $1$. A second pipe open at both ends is placed in the second chamber of gas density $2$. The compressibility of both the gases is equal.Calculate the length of the second pipe if the frequency of the first overtone in both the cases is equal.

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

  1. $\displaystyle \dfrac{4}{3}l _{1}\sqrt{\dfrac{\mathrm{p} _{2}}{\mathrm{p} _{1}}}$

  2. $\displaystyle \dfrac{4}{3}l _{1}\sqrt{\dfrac{\mathrm{p} _{1}}{\mathrm{p} _{2}}}$

  3. $l _{1}\sqrt{\dfrac{\mathrm{p} _{2}}{\mathrm{p} _{1}}}$

  4. $l _{1}\sqrt{\dfrac{\mathrm{p} _{1}}{\mathrm{p} _{2}}}$

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Correct Option: B
Explanation:

$l _{1}=\displaystyle \dfrac{3}{4}\dfrac{\mathrm{v} _{1}}{\mathrm{f} _{1}}$ , $l _{2}=\displaystyle \dfrac{\mathrm{v} _{2}}{\mathrm{f} _{2}}$

$\dfrac{3\mathrm{v} _{1}}{4l _{1}}=\dfrac{\mathrm{v} _{2}}{l _{2}}$

$l _{2}=\displaystyle \dfrac{4l _{1}\mathrm{v} _{2}}{3\mathrm{v} _{1}}=\dfrac{4l _{1}}{3}\sqrt{\dfrac{\mathrm{p} _{1}}{\mathrm{p} _{2}}}$

A steel wire of mass $4.0\ g$ and length $80\ cm$ is fixed at the two ends. The tension in the wire is $50\ N$. The wavelength of the fourth harmonic of the fundamental will be

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

  1. $80\ cm$

  2. $60\ cm$

  3. $40\ cm$

  4. $20\ cm$

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Correct Option: C
Explanation:

$m=$ mass per unit length

$=\cfrac { 4\times { 10 }^{ -3 } }{ 80\times { 10 }^{ -2 } } =0.005\quad Kg/m$

given $T=50 N$

$L=80 cm=0.8m$

$\therefore v=\sqrt { \cfrac { T }{ m }  } =\sqrt { \cfrac { 50 }{ 0.005 }  } =100m/sec$


Fundamental frequency 

${ f } _{ 0 }=\cfrac { 1 }{ 2L } \sqrt { \cfrac { T }{ m }  } \\ =\cfrac { 1 }{ 2\times 0.8 } \sqrt { \cfrac { 50 }{ 0.005 }  } \\ =625\quad Hz$


$\therefore { f } _{ 4 }$ Frequency of fourth harmonic 

$4{ f } _{ 0 }=4\times 62.5=250\quad Hz$

As we know

${ v } _{ 4 }={ f } _{ 4 }{ \lambda  } _{ 4 }\\ \therefore { \lambda  } _{ 4 }=\cfrac { { v } _{ 4 } }{ { f } _{ 4 } } =\cfrac { 100 }{ 250 } \\ =0.4m\quad \\ =40 cm$

The wave-function for a certain standing wave on a string fixed at both ends is $y\left( x,t \right) =0.5\sin { \left( 0.025\pi x \right)  } \cos { 500\ t } $ where $x$ and $y$ are in centimeters and t is in seconds. The shortest possible length of the string is: 

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

  1. $126\ cm$

  2. $160\ cm$

  3. $40\ cm$

  4. $80\ cm$

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Correct Option: C
Explanation:

For the shortest possible length, it should be allowing fundamental frequency resonance.

In fundamental frequency
$L=\dfrac { \lambda  }{ 2 } \quad \quad \quad \quad \left( \because K=\dfrac { 2\pi  }{ \lambda  }  \right) $
$=\dfrac { 2\pi /K }{ 2 } =\dfrac { \pi  }{ K } $
from $y=0.5\sin\left( 0.025\pi x \right) \cos\left( 500t \right) $
$K=0.025\pi \quad \quad \quad (on\quad comparing\quad with\quad y=A\sin\left( Kx \right) \cos\left( wt \right) )$
$\therefore \quad L=\dfrac { \pi  }{ 0.025\pi  } =\dfrac { 1000 }{ 25 } $
$\left[ L=40cm \right] $

Hence Option (C) is correct.

The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe 2 m long. The length of the open pipe is

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

  1. 8 m

  2. 4 m

  3. 2 m

  4. 1 m

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Correct Option: B
Explanation:

$l _0=$ length of organ pipe (open)

$l _c=$ length of close organ pipe
frequency of open pipe$=\cfrac{3V}{2l _0}$
frequency of close pipe $=\cfrac{3V}{2l _c}$
$\therefore \cfrac{3V}{2l _0}=\cfrac{3V}{4l _c}\2l _0=4l _c\2\times2m=4m$
B is the correct option.

A guitar string is $90 cm$ long and has a fundamental frequency of $124 Hz$. To produce a fundamental frequency of $186 Hz$, the guitar should be pressed at ?

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

  1. $60 cm$

  2. $30 cm$

  3. $20 cm$

  4. $ 10 cm$

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Correct Option: A
Explanation:

Here, $L _1 = 90 cm, \upsilon _1 = 124 Hz, \upsilon _2 = 186 Hz, L _2=?$

According to the law of length, $\upsilon _2L _2 = \upsilon _1L _1$

$\therefore \upsilon _2= \dfrac{\upsilon _1L _1}{\upsilon _2} = \dfrac{124 \times 90}{186} = 60 cm$