Tag: standing waves

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Questions Related to standing waves

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

The speed of mechanical waves depends on :-

  1. Density of medium

  2. Elasticity of medium

  3. Elasticity and density of medium

  4. Frequency of the wave.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The speed of a mechanical wave in a medium is determined by the medium's elastic properties (which provide the restoring force) and its inertial properties (density). Both factors are essential for wave propagation.

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

A suspension bridge is to be built across valley where it is known that the wind can gust at $5\ s$ intervals. It is estimated that the speed of transverse waves along the span of the bridge would be $400\ m/s$. The danger of reasonant motions in the bridge at its fundamental frequency would be greater if the span had a length of :

  1. $2000\ m$
  2. $1000\ m$
  3. $400\ m$
  4. $80\ m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Resonance occurs when the driving frequency matches the natural frequency of the system. For a bridge span, the fundamental frequency is f = v / (2L). Given v = 400 m/s and a period of 5s (frequency f = 0.2 Hz), setting 0.2 = 400 / (2L) leads to 0.4L = 400, so L = 1000 m.

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

A man generates a symmetrical pulse in a string by moving his hand up and down . At t = 0 the point in his hand moves downward. the pulse travels with speed of 3 m/s on the string & his hands passes 6 times in each second from the mean position. then the point on the string at a distance 3m will reach its upper extreme first time at times t =

  1. $1.25 sec$
  2. $1 sec$
  3. $\frac{{13}}{{12}}\sec $
  4. $0.25$secs
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The man's hand passes $6$ times from the mean position in $1$ sec, we can find that string creates $3$ cycles after $1$ sec.

Frequency of wave= $3Hz$
$V=f\lambda\Rightarrow \lambda=\cfrac {V}{f}=\cfrac {3}{3}=1m$
To reach upper extreme $\longrightarrow$ have to travel $3\lambda/4$ distance.
Time to travel $\cfrac {3 \lambda}{4}=\cfrac {3}{1}\times \cfrac {1}{3}=\cfrac {1}{4}=0.25$ sec

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

String 1 has twice the length, twice the radius, twice the tension and twice the density of another string 2. The relation between their fundamental frequencies of 1 and 2 is:

  1. $f _ { 1 } = 2 f _ { 2 }$
  2. $f _ { 1 } = 4 f _ { 2 }$
  3. $f _ { 2 } = 4 f _ { 1 }$
  4. $f _ { 1 } = f _ { 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The fundamental frequency of a stretched string is given by f = (1 / 2L) * sqrt(T / (pi * r^2 * d)). Substituting the doubled values for length L, radius r, tension T, and density d gives f1 = (1 / 2(2L)) * sqrt(2T / (pi * (2r)^2 * (2d))) = f2 / 4, which means f2 = 4 f1.

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

In a reasonance tube experiment, a closed organ pipe of lenght $120$ cm is used. initially it is completely fiiled with water. It is vibrated with tuning fork of frequency $340$ Hz. To achieve reasonance the water level is lowered then (given ${V _{air}} = 340m/\sec $., neglect end correction):

  1. minimum lenght of water column to have the resonance is 45 cm.

  2. the distance between two successive nodes is 50 cm.

  3. the maximum lenght of water column to resonance is 95 cm.

  4. the distance between two successive nodes is 25 cm.
Reveal answer Fill a bubble to check yourself
D Correct answer
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.