Tag: standing waves in strings

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Questions Related to standing waves in strings

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

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

  1. $2\, V _o$
  2. $0.5$
  3. $2.6$
  4. $1.6$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Frequency f is proportional to (1 / L) * sqrt(T). If L is reduced by 35%, L_new = 0.65 L_old. If T is increased by 69%, T_new = 1.69 T_old. f_new / f_old = (L_old / L_new) * sqrt(T_new / T_old) = (1 / 0.65) * sqrt(1.69) = 1.3 / 0.65 = 2. Thus, f_new = 2 * f_old.

Multiple choice physics oscillations and waves standing waves in strings standing waves 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:

  1. double

  2. equal

  3. four times

  4. eight times
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $

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

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

  1. 96 Hz

  2. 192 Hz

  3. 1200 Hz

  4. 2400 Hz
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

The vibrating body while playing a violin is ___________.

  1. wire

  2. the box of the violin

  3. both wire and box

  4. only air
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a violin, the vibrating string (wire) creates the initial sound, but the wooden body of the violin acts as a resonator to amplify the sound waves, making both essential for the instrument's function.

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

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.

  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. $l _{1}\sqrt{\dfrac{\mathrm{p} _{1}}{\mathrm{p} _{2}}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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

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

  1. $80\ cm$
  2. $60\ cm$
  3. $40\ cm$
  4. $20\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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

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: 

  1. $126\ cm$
  2. $160\ cm$
  3. $40\ cm$
  4. $80\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.