Tag: standing waves in strings

Questions Related to standing waves in strings

First overtone frequency of a closed pipe of length $l _1$ is equal to the$^{2nd}$ Harmonic frequency of an open pipe of length $l _2$. The ratio $l _1 \, l _2.$

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

  1. $3/4$

  2. $4/3$

  3. $3/2$

  4. $2/3$

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Correct Option: C

The fundamental frequency of a stretched string is $V _o$. If the length is reduced by $35$% and tension increased by $69$% the fundamental frequency will be

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

  1. $2\, V _o$

  2. $0.5$

  3. $2.6$

  4. $1.6$

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Correct Option: A

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be: (Assume that the highest frequency a person can hear is 20.000Hz)

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

  1. 6

  2. 4

  3. 5

  4. 7

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Correct Option: D

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.