JEE Advanced PYQ Vibration of String and Organ Pipe

Q 1 :

A student is performing the experiment of resonance column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is $38 ° C$ in which the speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is [2012]

14.0 cm
15.2 cm
16.4 cm
17.6 cm

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(2)

$Considering the end correction$ $[e = 0.3 D where D = diameter]$

$f = n [\frac{v}{4 (l + e)}]$ $For first resonance, n = 1$

$∴$ $f = \frac{v}{4 (l + 0.3 D)} \Rightarrow l = \frac{v}{4 f} - 0.3 D$

$or, l = (\frac{336 \times 100}{4 \times 512}) - 0.3 \times 4 = 15.2 cm$

Q 2 :

An open pipe is in resonance in 2nd harmonic with frequency $f_{1}$ . Now one end of the tube is closed and frequency is increased to $f_{2}$ such that the resonance again occurs in $n$ th harmonic. Choose the correct option. [2005]

$n = 3, f_{2} = \frac{3}{4} f_{1}$
$n = 3, f_{2} = \frac{5}{4} f_{1}$
$n = 5, f_{2} = \frac{3}{4} f_{1}$
$n = 5, f_{2} = \frac{5}{4} f_{1}$

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(4)

$Frequency of 2nd harmonic of open pipe,$

$f_{1} = \frac{v}{λ} = \frac{v}{l} \dots (i)$

$Frequency of n^{th} harmonic of closed pipe,$

$f_{2} = \frac{v}{λ} = \frac{n v}{4 l} \dots (i i)$

$Here n is an odd number. From eq. (i) and (ii)$

$f_{2} = \frac{n}{4} f_{1},$ $For first resonance, n = 5$ $∴ f_{2} = \frac{5}{4} f_{1}$

Q 3 :

A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ${ms}^{- 1}$ , the mass of the string is [2010]

5 grams
10 grams
20 grams
40 grams

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(2)

$Frequency of 2nd harmonic of string = fundamental frequency produced in the pipe$

$2 (\frac{v_{1}}{2 l_{1}}) = \frac{v_{2}}{4 l_{2}}$

$∴$ $2 \times [\frac{1}{2 l_{1}} \sqrt{\frac{T}{μ}}] = \frac{v}{4 l_{2}}$

$∴$ $\frac{1}{0.5} \sqrt{\frac{50}{μ}} = \frac{320}{4 \times 0.8}$ $\Rightarrow μ = 0.02 {kg m}^{- 1}$

Hence, the mass of the string, $m_{1} = μ l_{1}$

$= 0.02 \times 0.5 kg = 10 g$

Q 4 :

In the experiment to determine the speed of sound using a resonance column, [2007]

prongs of the tuning fork are kept in a vertical plane
prongs of the tuning fork are kept in a horizontal plane
in one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air
in one of the two resonances observed, the length of the resonating air column is close to half of the wavelength of sound in air

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(1)

To determine the speed of sound using a resonance column, prongs of the tuning fork are kept in a vertical plane. As shown in the figure, the fringes of the tuning fork are kept in a vertical plane.

Q 5 :

A massless rod of length $L$ is suspended by two identical strings $A B$ and $C D$ of equal length. A block of mass $m$ is suspended from point $O$ such that $B O$ is equal to $x$ . Further it is observed that the frequency of 1st harmonic in $A B$ is equal to 2nd harmonic frequency in $C D$ . $‘ x ’$ is [2006]

$\frac{L}{5}$
$\frac{4 L}{5}$
$\frac{3 L}{4}$
$\frac{L}{4}$

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(1)

$Frequency of first harmonic in AB = f_{1} = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$

$Frequency of second harmonic in CD = f_{2} = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$

$f_{1} = f_{2} (given)$

$∴$ $\frac{1}{2 l} \sqrt{\frac{T_{1}}{μ}} = \frac{1}{l} \sqrt{\frac{T_{2}}{μ}}$ $or$ $T_{1} = 4 T_{2} . . . (i)$

Equating torques due to $T_{1}$ and $T_{2}$ about $O$ for rotational equilibrium,

$∴$ $T_{1} x = T_{2} (L - x)$

For translational equilibrium,

$T_{1} + T_{2} = m g$

From eq. (i) and (iii),

$T_{1} = \frac{4 m g}{5} and T_{2} = \frac{m g}{5}$

Now from eq. (ii),

$\frac{4 m g}{5} \times x = \frac{m g}{5} (L - x)$ $\Rightarrow 4 x = L - x$

$∴$ $x = \frac{L}{5}$

Q 6 :

In a resonance tube with tuning fork of frequency 512 Hz, first resonance occurs at water level equal to 30.3 cm and second resonance occurs at 63.7 cm. The maximum possible error in the speed of sound is [2005]

51.2 cm/s
102.4 cm/s
204.8 cm/s
153.6 cm/s

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(3)

In a resonance tube,

$ℓ_{1} + e = \frac{λ}{4}$ $ℓ_{2} + e = \frac{3 λ}{4}$

But $v = ν λ$

$∴$ $v = ν \frac{4}{3} (ℓ_{2} + e)$ $\Rightarrow ℓ_{2} + e = \frac{3 v}{4 ν} . . . (i)$

$∴$ $v = ν 4 (ℓ_{1} + e)$ $\Rightarrow ℓ_{1} + e = \frac{v}{4 ν} . . . (i i)$

Subtracting (i) and (ii),

$v = 2 ν (ℓ_{2} - ℓ_{1})$

$∴$ $Δ v = 2 ν (Δ ℓ_{2} + Δ ℓ_{1})$

$= 2 \times 512 \times (0.1 + 0.1) cm/s$

$= 204.8 cm/s$

Hence, maximum possible error in speed, $Δ v = 204.8 cm/s$

Q 7 :

A pipe of length $ℓ_{1}$ , closed at one end is kept in a chamber of gas of density $ρ_{1}$ . A second pipe open at both ends is placed in a second chamber of gas of density $ρ_{2}$ . The compressibility of both the gases is equal. Calculate the length of the second pipe if frequency of first overtone in both the cases is equal. [2004]

$\frac{4}{3} ℓ_{1} \sqrt{\frac{ρ_{2}}{ρ_{1}}}$
$\frac{4}{3} ℓ_{1} \sqrt{\frac{ρ_{1}}{ρ_{2}}}$
$ℓ_{1} \sqrt{\frac{ρ_{2}}{ρ_{1}}}$
$ℓ_{1} \sqrt{\frac{ρ_{1}}{ρ_{2}}}$

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(2)

Frequency of first overtone in closed organ pipe,

$ν = \frac{3 v}{4 ℓ_{1}} \sqrt{\frac{P}{ρ_{1}}}$

Frequency of first overtone in open organ pipe,

$ν' = \frac{1}{ℓ_{2}} \sqrt{\frac{P}{ρ_{2}}}$

Here, $ν = ν'$

$\frac{3 v}{4 ℓ_{1}} \sqrt{\frac{P}{ρ_{1}}} = \frac{1}{ℓ_{2}} \sqrt{\frac{P}{ρ_{2}}}$

$∴$ $ℓ_{2} = \frac{4}{3} ℓ_{1} \sqrt{\frac{ρ_{1}}{ρ_{2}}}$

Q 8 :

A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is [2002]

25 kg
5 kg
12.5 kg
1/25 kg

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(1)

$Fundamental frequency, f_{0} = \frac{P}{2 ℓ} \sqrt{\frac{T}{μ}}$

$= \frac{5}{2 ℓ} \sqrt{\frac{9 g}{μ}} = \frac{3}{2 ℓ} \sqrt{\frac{M g}{μ}}$

$\Rightarrow M = 25 kg$

As frequency corresponds to 5th and 3rd harmonic as $P = 5$ and $P = 3$ respectively.

Q 9 :

Two vibrating strings of the same material but lengths $L$ and $2 L$ have radii $2 r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length $L$ with frequency $ν_{1}$ and the other with frequency $ν_{2}$ . The ratio $ν_{1} / ν_{2}$ is given by [2000]

2
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8
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(4)

$n_{1} = \frac{1}{2 ℓ} \sqrt{(\frac{T}{4 π r^{2} ρ})}$ $and$ $n_{2} = \frac{1}{4 ℓ} \sqrt{(\frac{T}{π r^{2} ρ})}$

$n = \frac{v}{λ} = \frac{1}{λ} \sqrt{\frac{T}{m}}$ $[where \frac{λ}{2} = length of string]$

$∴$ $\frac{n_{1}}{n_{2}} = 2 \times \frac{1}{2} = 1$ $[∵ m = \frac{mass}{length} = \frac{ρ \times A \times length}{length} = ρ A]$

Q 10 :

A string of length $1 m$ and mass $2 \times 10^{- 5} kg$ is under tension $T$ . When the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is ________ Newton. [2023]

(5)

As two successive harmonics are found to occur at frequency 750 Hz and 1000 Hz

$f = \frac{P}{2 ℓ} \sqrt{\frac{T}{μ}}$

So, $750 = \frac{P}{2} \sqrt{\frac{T}{μ}} . . . (i)$

and, $1000 = \frac{P + 1}{2} \sqrt{\frac{T}{μ}} . . . (i i)$

Dividing eq. (ii) by (i),

$\frac{4}{3} = \frac{P + 1}{P}$

$∴$ $P = 3$

Putting this value of $P = 3$ in eq. (ii) and solving we get tension $T$ ,

$1000 = \frac{4}{2} \sqrt{\frac{T}{2 \times 10^{- 5}}}$

$∴$ $T = 5 N$

Q 11 :

A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string. [2009]

(5)

The distance between two successive nodes

$D = \frac{λ}{2} = \frac{v}{2 f} = \frac{\sqrt{T / μ}}{2 f} = \frac{\sqrt{\frac{0.5 \times 0.2}{10^{- 3}}}}{2 \times 100} = \frac{10}{2} = 5 cm$

Q 12 :

A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of 2 ${ms}^{- 1}$ in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is 320 ${ms}^{- 1}$ , the smallest value of the percentage change required in the length of the pipe is [2020]

(0.62)

Let $ℓ_{1} = initial length of pipe$

$ℓ_{2} = new length of pipe$

$V_{T} = Speed of tuning fork$

In closed organ pipe, $f = \frac{V}{4 ℓ_{1}}$

When tuning fork is moved, $f' = f (\frac{V}{V - V_{T}}) = \frac{V}{4 ℓ_{2}}$

$\Rightarrow \frac{V}{4 ℓ_{1}} (\frac{V}{V - V_{T}}) = \frac{V}{4 ℓ_{2}}$ $\Rightarrow \frac{V - V_{T}}{V} = \frac{ℓ_{2}}{ℓ_{1}}$

$\Rightarrow \frac{ℓ_{2}}{ℓ_{1}} - 1 = \frac{V - V_{T}}{V} - 1$ $\Rightarrow \frac{ℓ_{2} - ℓ_{1}}{ℓ_{1}} = - \frac{V_{T}}{V}$

Percentage change required in the length of the pipe,

$\frac{ℓ_{2} - ℓ_{1}}{ℓ_{1}} \times 100 = \frac{- 2}{320} \times 100 = - 0.625 %$

Hence, smallest value of percentage change required in the length of pipe is 0.625%

Q 13 :

Consider a system of three connected strings, $S_{1}$ , $S_{2}$ and $S_{3}$ with uniform linear mass densities $μ kg/m$ , $4 μ kg/m$ and $16 μ kg/m$ , respectively, as shown in the figure. $S_{1}$ and $S_{2}$ are connected at the point $P$ , whereas $S_{2}$ and $S_{3}$ are connected at the point $Q$ , and the other end of $S_{3}$ is connected to a wall. A wave generator $O$ is connected to the free end of $S_{1}$ . The wave from the generator is represented by $y = y_{0} \cos (ω t - k x) cm,$ where $y_{0}$ , $ω$ and $k$ are constants of appropriate dimensions. Which of the following statements is/are correct: [2025]

When the wave reflects from $P$ for the first time, the reflected wave is represented by $y = α_{1} y_{0} \cos (ω t + k x + π) cm,$ where $α_{1}$ is a positive constant.
When the wave transmits through $P$ for the first time, the transmitted wave is represented by $y = α_{2} y_{0} \cos (ω t - k x) cm,$ where $α_{2}$ is a positive constant.
When the wave reflects from $Q$ for the first time, the reflected wave is represented by $y = α_{3} y_{0} \cos (ω t - k x + π) cm,$ where $α_{3}$ is a positive constant.
When the wave transmits through $Q$ for the first time, the transmitted wave is represented by $y = α_{4} y_{0} \cos (ω t - 4 k x) cm,$ where $α_{4}$ is a positive constant.

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Select one or more options

(1, 4)

When wave going from rarer to denser, phase change by $π$

$y_{i} = y_{0} \cos (ω t - k x)$

$y_{r} = A_{r} \cos (ω t + k x + π)$

$y_{r} = α_{1} y_{0} \cos (ω t + k x + π)$

So, option (1) correct.

When transmitted from point P,

$y_{t} = A_{t} \cos [ω t - k_{1} x]$

$\frac{k_{1}}{k} = \sqrt{\frac{μ_{1}}{μ}} = \frac{k_{1}}{k} = \sqrt{\frac{4 μ}{μ}}$

$k_{1} = 2 k$

$y_{t} = α_{2} y_{0} \cos [ω t - 2 k x]$

So, option (2) incorrect.

When reflected from $Q$ ,

$y_{i} = a_{2} y_{0} \cos [ω t - 2 k x]$

$y_{r} = a_{3} y_{0} \cos [ω t + 2 k x + π]$

So, option (3) incorrect.

When transmitted from $Q$ ,

$y_{t} = α_{4} y_{0} \cos [ω t - k_{2} x]$

$\frac{k_{2}}{2 k} = \sqrt{\frac{16 μ}{4 μ}}$

$k_{2} = 4 k$

$y_{t} = α_{4} y_{0} \cos [ω t - 4 k x]$

So, option (4) correct.

Q 14 :

Two uniform strings of mass per unit length $μ$ and $4 μ$ , and length $L$ and $2 L$ , respectively, are joined at point $O$ , and tied at two fixed ends $P$ and $Q$ , as shown in the figure. The strings are under a uniform tension $T$ . If we define the frequency $ν_{0} = \frac{1}{2 L} \sqrt{\frac{T}{μ}},$ which of the following statement(s) is(are) correct? [2024]

With a node at $O$ , the minimum frequency of vibration of the composite string is $ν_{0}$ .
With an antinode at $O$ , the minimum frequency of vibration of the composite string is $2 ν_{0}$ .
When the composite string vibrates at the minimum frequency with a node at $O$ , it has 6 nodes, including the end nodes.
No vibrational mode with an antinode at $O$ is possible for the composite string.

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Select one or more options

(1, 3, 4)

The velocity of a transverse wave in a stretched string, $C = \sqrt{\frac{T}{μ}}$

$C_{1} = \sqrt{\frac{T}{μ}}, C_{2} = \sqrt{\frac{T}{4 μ}} = \frac{C_{1}}{2}$

For node at $O$ ,

$L = \frac{n λ_{1}}{2} and 2 L = \frac{m λ_{2}}{2}$ $(Here, n, m are integers)$

or, $λ_{1} = \frac{2 L}{n} and λ_{2} = \frac{4 L}{m}$

$\frac{C_{1}}{λ_{1}} = \frac{C_{2}}{λ_{2}}$ $\Rightarrow \frac{C_{1}}{\frac{2 L}{n}} = \frac{\frac{C_{1}}{2}}{\frac{4 L}{m}}$

$∴$ $4 n = m$

For minimum frequency, $n = 1, m = 4$

$∴$ $ν_{\min} = \frac{C_{1} \times 1}{2 L} = \frac{1}{2 L} \sqrt{\frac{T}{μ}} = ν_{0}$

So, option (1) is correct.

The string will look like

i.e. 6 nodes including the end nodes so option (3) is correct.

For antinode at $O$ ,

$L = \frac{(2 n + 1) λ_{1}}{4} and 2 L = \frac{(2 m + 1) λ_{2}}{4}$ $(n, m are integers)$

or, $λ_{1} = \frac{4 L}{(2 n + 1)} and λ_{2} = \frac{8 L}{(2 m + 1)}$

$m \cdot \frac{1}{4 L} \sqrt{\frac{T}{μ}} = n \cdot \frac{1}{4 (2 L)} \sqrt{\frac{T}{4 μ}}$

$\Rightarrow \frac{m}{n} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$∴$ $m = 1, f_{\min} = 1 \cdot \frac{1}{4 L} \sqrt{\frac{T}{μ}} = \frac{ν_{0}}{2}$

So, option (2) is incorrect.

Q 15 :

In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500 Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7 cm and 83.9 cm. Which of the following statements is (are) true? [2017]

The speed of sound determined from this experiment is 332 ${ms}^{- 1}$
The end correction in this experiment is 0.9 cm
The wavelength of the sound wave is 66.4 cm
The resonance at 50.7 cm corresponds to the fundamental harmonic

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Select one or more options

(1, 2, 3)

According to question, the length of the air column is varied by changing the level of water in the resonance tube,

so, $(2 n + 1) \frac{λ}{4} = 50.7 + e . . . (i)$

and $(2 n + 3) \frac{λ}{4} = 83.9 + e . . . (i i)$

Dividing eq. (i) by (ii)

If $n = 1$ , $\frac{3 λ / 4}{5 λ / 4} = \frac{50.7 + e}{83.9 + e}$

$∴$ $3 \times 83.9 + 3 e = 5 \times 50.7 + 5 e$

$\Rightarrow 2 e = 1.8$ $∴$ $e = 0.9 cm$

$∴$ $\frac{3 λ}{4} = 50.7 + 0.9 = 51.6$

$\Rightarrow λ = 66.4 cm = 0.664 m$

Also speed of sound, $V = ν λ = 500 \times 0.664 {ms}^{- 1} = 332.0 {ms}^{- 1}$

Q 16 :

One end of a taut string of length 3 m along the $x$ -axis is fixed at $x = 0$ . The speed of the waves in the string is $100 {ms}^{- 1}$ . The other end of the string is vibrating in the $y$ -direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is(are) [2014]

$y (t) = A \sin (\frac{π x}{6}) \cos (\frac{50 π t}{3})$
$y (t) = A \sin (\frac{π x}{3}) \cos (\frac{100 π t}{3})$
$y (t) = A \sin (\frac{5 π x}{6}) \cos (\frac{250 π t}{3})$
$y (t) = A \sin (\frac{5 π x}{2}) \cos (250 π t)$

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Select one or more options

(1, 3, 4)

There should be a displacement node at $x = 0$ and a displacement antinode at $x = 3 m$ .

Therefore, $y = 0$ at $x = 0$ and $y = \pm A$ at $x = 3 m$ .

Speed of wave, $v = \frac{ω}{k} = 100 {ms}^{- 1}$

Hence options (1), (3) & (4) satisfy the above conditions.

Q 17 :

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation,

$y (x, t) = (0.01 m) \sin [(62.8 m^{- 1}) x] \cos [(628 s^{- 1}) t] .$ Assuming $π = 3.14$ , the correct statement(s) is (are) [2013]

The number of nodes is 5
The length of the string is 0.25 m
The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m
The fundamental frequency is 100 Hz

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Select one or more options

(2, 3)

$y = [0.01 \sin (62.8 x)] \cos (628 t)$ $[Given]$

$From the given equation, k = \frac{2 π}{λ} = 62.8$

$∴$ $λ = \frac{2 π}{62.8} = 0.1 m$

Length of string, $l = 5 \times \frac{λ}{2} = 5 \times \frac{1}{20} = 0.25 m$

The midpoint $M$ is an antinode and has the maximum displacement $= 0.01 m$

The fundamental frequency, $ν = \frac{v}{2 l} = \frac{ω / k}{2 l}$

$= \frac{628}{2 \times 0.25 \times 62.8} = 20 Hz$

Q 18 :

A musical instrument is made using four different metal strings 1, 2, 3 and 4 with mass per unit length $μ$ , $2 μ$ , $3 μ$ and $4 μ$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_{0}$ and $2 L_{0}$ . It is found that in string-1 $(μ)$ at free length $L_{0}$ and tension $T_{0}$ the fundamental mode frequency is $f_{0}$ .

List-I gives the above four strings while List-II lists the magnitude of some quantity. [2019]

List-I List-II

(I) String-1 $(μ)$ (P) 1

(II) String-2 $(2 μ)$ (Q) $\frac{1}{2}$

(III) String-3 $(3 μ)$ (R) $\frac{1}{\sqrt{2}}$

(IV) String-4 $(4 μ)$ (S) $\frac{1}{\sqrt{3}}$

(T) $\frac{3}{16}$

(U) $\frac{1}{16}$

If the tension in each string is $T_{0}$ , the correct match for the highest fundamental frequency in $f_{0}$ units will be,

I → Q, II → P, III → R, IV → T
I → Q, II → S, III → R, IV → P
I → P, II → R, III → S, IV → Q
I → P, II → Q, III → T, IV → S

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(3)

$Frequency, ν = \frac{1}{2 ℓ} \sqrt{\frac{T}{m}} for first mode of vibration$

For $' ν'$ to be maximum, $' ℓ'$ should be minimum.

$String-1 f_{0} = \frac{1}{2 L_{0}} \sqrt{\frac{T_{0}}{μ}}$

$String-2 f_{2} = \frac{1}{2 L_{0}} \sqrt{\frac{T_{0}}{2 μ}} = \frac{f_{0}}{\sqrt{2}}$

$String-3 f_{3} = \frac{1}{2 L_{0}} \sqrt{\frac{T_{0}}{4 μ}} = \frac{f_{0}}{\sqrt{3}}$

$String-4 f_{4} = \frac{1}{2 L_{0}} \sqrt{\frac{T_{0}}{4 μ}} = \frac{f_{0}}{2}$

Q 19 :

A musical instrument is made using four different metal strings 1, 2, 3 and 4 with mass per unit length $μ$ , $2 μ$ , $3 μ$ and $4 μ$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_{0}$ and $2 L_{0}$ . It is found that in string-1 $(μ)$ at free length $L_{0}$ and tension $T_{0}$ , the fundamental mode frequency is $f_{0}$ .

List-I gives the above four strings while List-II lists the magnitude of some quantity. [2019]

List-I List-II

(I) String-1 $(μ)$ (P) 1

(II) String-2 $(2 μ)$ (Q) $\frac{1}{2}$

(III) String-3 $(3 μ)$ (R) $\frac{1}{\sqrt{2}}$

(IV) String-4 $(4 μ)$ (S) $\frac{1}{\sqrt{3}}$

(T) $\frac{3}{16}$

(U) $\frac{1}{16}$

The lengths of the strings 1, 2, 3 and 4 are kept fixed at $L_{0}$ , $\frac{3 L_{0}}{2}$ , $\frac{5 L_{0}}{4}$ , and $\frac{7 L_{0}}{4}$ , respectively. Strings 1, 2, 3, and 4 are vibrated at their $1^{st}$ , $3^{rd}$ , $5^{th}$ , and $14^{th}$ harmonics, respectively such that all the strings have same frequency.

The correct match for the tension in the four strings in the units of $T_{0}$ will be

I → T, II → Q, III → R, IV → U
I → P, II → Q, III → T, IV → U
I → P, II → Q, III → R, IV → T
I → P, II → R, III → T, IV → U

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4

(2)

$As ν = \frac{p}{2 ℓ} \sqrt{\frac{T}{m}}$

$∴$ $T = \frac{ν^{2} ℓ^{2} m}{p^{2}}$

$String-1 T_{0} = \frac{f_{0}^{2} 4 L_{0}^{2} μ}{ℓ^{2}}$

$String-2 T_{2} = \frac{f_{0}^{2} 4 {(\frac{3}{2})}^{2} L_{0}^{2} (2 μ)}{{(3)}^{2}} = \frac{T_{0}}{2}$

$String-3 T_{3} = \frac{f_{0}^{2} 4 {(\frac{5}{2})}^{2} L_{0}^{2} (3 μ)}{5^{2}} = \frac{3}{16} T_{0}$

$String-4 T_{4} = \frac{f_{0}^{2} 4 {(\frac{7}{4})}^{2} L_{0}^{2} (4 μ)}{{(14)}^{2}} = \frac{T_{0}}{16}$

Q 20 :

Column I shows four systems, each of the same length $L$ , for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as $λ_{f}$ . Match each system with statements given in Column II describing the nature and wavelength of the standing waves. [2011]

Column I Column II

(A)
Pipe closed at one end

(p) Longitudinal waves

(B)
Pipe open at both ends

(q) Transverse waves

(C)
Stretched wire clamped at both ends

(r) $λ_{f} = L$

(D)
Stretched wire clamped at both ends and at mid-point

(s) $λ_{f} = 2 L$

(t) $λ_{f} = 4 L$

$A - p, t; B - p, s; C - q, s; D - q, r$
$A - q, r; B - p, s; C - q, s; D - p, t$
$A - q, r; B - q, s; C - p, s; D - p, t$
$A - p, s; B - q, s; C - q, r; D - p, t$

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(1)

(A) Pipe closed at one end

Waves produced are longitudinal

$\frac{λ_{f}}{4} = L$

$∴$ $λ_{f} = 4 L$

(B) Pipe open at both ends

waves produced are longitudinal

$\frac{λ_{f}}{2} = L$

$∴$ $λ_{f} = 2 L$

(C) Stretched wire clamped at both ends

Waves produced are transverse in nature.

$\frac{λ_{f}}{2} = L$

$∴$ $λ_{f} = 2 L$

(D) Stretched wave clamped at both ends and at mid point

Waves produced are transverse in nature

$λ_{f} = L$

Topic Question Set

Q 2 :

An open pipe is in resonance in 2nd harmonic with frequency $f_{1}$ . Now one end of the tube is closed and frequency is increased to $f_{2}$ such that the resonance again occurs in $n$ th harmonic. Choose the correct option. [2005]

Q 4 :

In the experiment to determine the speed of sound using a resonance column, [2007]

Q 6 :

In a resonance tube with tuning fork of frequency 512 Hz, first resonance occurs at water level equal to 30.3 cm and second resonance occurs at 63.7 cm. The maximum possible error in the speed of sound is [2005]

Q 10 :

A string of length $1 m$ and mass $2 \times 10^{- 5} kg$ is under tension $T$ . When the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is ________ Newton. [2023]

Q 11 :

A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string. [2009]

Q 17 :

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation,

$y (x, t) = (0.01 m) \sin [(62.8 m^{- 1}) x] \cos [(628 s^{- 1}) t] .$ Assuming $π = 3.14$ , the correct statement(s) is (are) [2013]

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	List-I		List-II
(I)	String-1 $(μ)$	(P)	1
(II)	String-2 $(2 μ)$	(Q)	$\frac{1}{2}$
(III)	String-3 $(3 μ)$	(R)	$\frac{1}{\sqrt{2}}$
(IV)	String-4 $(4 μ)$	(S)	$\frac{1}{\sqrt{3}}$
		(T)	$\frac{3}{16}$
		(U)	$\frac{1}{16}$

	Column I		Column II
(A)	Pipe closed at one end	(p)	Longitudinal waves
(B)	Pipe open at both ends	(q)	Transverse waves
(C)	Stretched wire clamped at both ends	(r)	$λ_{f} = L$
(D)	Stretched wire clamped at both ends and at mid-point	(s)	$λ_{f} = 2 L$
		(t)	$λ_{f} = 4 L$

Topic Question Set

Q 2 : An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closed and frequency is increased to f2 such that the resonance again occurs in nth harmonic. Choose the correct option. [2005]

Q 4 : In the experiment to determine the speed of sound using a resonance column, [2007]

Q 5 : A massless rod of length L is suspended by two identical strings AB and CD of equal length. A block of mass m is suspended from point O such that BO is equal to x. Further it is observed that the frequency of 1st harmonic in AB is equal to 2nd harmonic frequency in CD. ‘x’ is [2006]

Q 6 : In a resonance tube with tuning fork of frequency 512 Hz, first resonance occurs at water level equal to 30.3 cm and second resonance occurs at 63.7 cm. The maximum possible error in the speed of sound is [2005]

Q 10 : A string of length 1 m and mass 2×10-5 kg is under tension T. When the string vibrates, two successive harmonics are found to occur at frequencies 750 Hz and 1000 Hz. The value of tension T is ________ Newton. [2023]

Q 11 : A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string. [2009]

Q 17 : A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x,t)=(0.01 m)sin[(62.8 m-1)x]cos[(628 s-1)t]. Assuming π=3.14, the correct statement(s) is (are) [2013]

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Q 2 :

An open pipe is in resonance in 2nd harmonic with frequency $f_{1}$ . Now one end of the tube is closed and frequency is increased to $f_{2}$ such that the resonance again occurs in $n$ th harmonic. Choose the correct option. [2005]

Q 4 :

In the experiment to determine the speed of sound using a resonance column, [2007]

Q 6 :

In a resonance tube with tuning fork of frequency 512 Hz, first resonance occurs at water level equal to 30.3 cm and second resonance occurs at 63.7 cm. The maximum possible error in the speed of sound is [2005]

Q 10 :

A string of length $1 m$ and mass $2 \times 10^{- 5} kg$ is under tension $T$ . When the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is ________ Newton. [2023]

Q 11 :

A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string. [2009]

Q 17 :

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation,

$y (x, t) = (0.01 m) \sin [(62.8 m^{- 1}) x] \cos [(628 s^{- 1}) t] .$ Assuming $π = 3.14$ , the correct statement(s) is (are) [2013]