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

A string of length $1m$ and linear mass density $0.01kgm^{-1}$ is stretched to a tension of $100N$. When both ends of the string are fixed, the three lowest frequencies for standing wave are $f _{1}, f _{2}$ and $f _{3}$. When only one end of the string is fixed, the three lowest frequencies for standing wave are $n _{1}, n _{2}$ and $n _{3}$. Then 

  1. $n _{3} = 5n _{1} = f _{3} = 125 Hz $
  2. $f _{3} = 5f _{1} = n _{2} = 125 Hz $
  3. $f _{3} = n _{2} = 3f _{1} = 150 Hz $
  4. $n _{2} = \displaystyle \dfrac {f _{1} + f _{2}}{2} = 75 Hz $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When both ends are fixed, the string forms a length half the wavelength. That is, it has two nodes at the ends. For the next frequency, it will have the length equals the wavelength. So, the general formula for length of the string becomes $L = n \lambda /2$.


For the string fixed on only one end, there is always an anti node at one end and a node at the other end. So, the length of the string gets divided into $1/4th$ of the wavelength ($\lambda$). The general formula for the length of the string is $L' = n \lambda /4.$

The frequency $f$ becomes $V/ \lambda$, $V$ is the velocity. In the first case, frequency $f = nV/2L,$   $n = 1,2,3,....$

In the second case, it is $nV/4L$,    $n = 1,3,5,7......$ because of the length of the string will always have a half wave present. This makes n an odd number.
For the first case: 
$f _1 = 1/2L(\sqrt{(T/ \mu)}) = 50 Hz = V/2 \times L$
$f _2 = 2\times f _1 = 100 Hz = V/L$
$f3 = 3\times f _1 = 150 Hz = 3V/2\times L$

Second case:
$n _1 = V/4L$
$n _2 = 3V/4L = (f-1+f _2)/2 = (100+50)/2 = 75 Hz$

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

A massless rod of length $l$  is hung from the ceiling with the help of two identical wires attached at its ends. A block is hung on the rod at a distance $x$ from the left end. In the case, the frequency of the $1st$ harmonic of the wire on the left end is equal to the frequency of the $2nd$ harmonic of the wire on the right. The value of $x$ is

  1. $\displaystyle \dfrac{l}{2}$
  2. $\displaystyle \dfrac{l}{3}$
  3. $\displaystyle \dfrac{l}{4}$
  4. $\displaystyle \dfrac{l}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since, the frequency of the first harmonic from the left is equal to that of second harmonic from right,
$ {\nu} _{1} = 2{\nu} _{2} $
Hence, $ {T} _{1} = {T} _{2} $
Thus, according to the question,
$ {T} _{1} (x) = {T} _{2} (l - x) $
Solving this equation for $ {T} _{1}$ and ${T} _{2} $ we get the value of x = $ \dfrac{l}{5} $

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

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

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

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