Guitar string of length 1.5 m is made of steel of density 7800 kg/m³ and Young’s modulus [latex]2\times 10^{11}[/latex] N/m². It produces an elastic strain of 1% in the string. Calculate.
i. stress developed in the string.
ii. frequency of second mode of vibration.

Solution
Length of string [latex](l)[/latex] = 1.5 m
Density [latex](\rho)[/latex] = 7800 kg/m³
Young’s modulus (Y) = [latex]2\times 10^{11}[/latex] N/m²
Strain (%) = 1% = [latex]\frac{1}{100}[/latex] = 0.01
i.    Stress developed (T/A) =?
We know,
[latex]Y = \frac{Stress}{Strain}[/latex]
Or, [latex]2\times 10^{11} = \frac{Stress}{0.01}[/latex]
Or, Stress (T/A) = [latex]2\times 10^9[/latex] N/m²
ii.   frequency of second mode of vibration ([latex]f_2[/latex]) =?
We know,
Fundamental frequency [latex](f_1) = \frac{1}{2l}\sqrt{\frac{T}{\mu}}[/latex]
     = [latex]\frac{1}{2l}\sqrt{\frac{T}{m/l}}[/latex]
    = [latex]\frac{1}{2l}\sqrt{\frac{Tl}{\rho V}}[/latex]        [[latex]\because m = \rho . V[/latex]]
    = [latex]\frac{1}{2l}\sqrt{\frac{Tl}{\rho Al}} = \frac{1}{2l}\sqrt{\frac{T}{\rho A}}[/latex]
  = [latex]\frac{1}{2l}\sqrt{\frac{Stress}{\rho}}[/latex] [[latex]\because Stress = \frac{T}{A}[/latex]]
  = [latex]\frac{1}{2\times 1.5}\sqrt{\frac{2\times 10^9}{7800}}[/latex]
= 168.79 Hz
So, [latex]f_2 = 2\times f_1[/latex]
= [latex] 2\times 168.79[/latex]            
= 337.58 Hz

A sonometer wire of density 9 [latex]gm/cm^3[/latex] and length 1m is subjected to an extension of 0.05 cm. What is the lowest frequency of transverse vibration in the wire? Assume Young’s modulus, [latex]Y = 9\times 10^{10}\ N/m^2[/latex].

Solution
Density of wire [latex](\rho)[/latex] = 9 gm/cm³
= [latex]9\times 10^3\ kg/m^3[/latex]
Length of wire [latex](l)[/latex] = 1m
Extension (e) = 0.05 cm = [latex]0.05\times 10^{-2}[/latex] m = [latex](5\times 10^{-4})[/latex] m
Fundamental frequency ([latex]f_1[/latex]) =?
Young’s modulus [latex](Y) = 9\times 10^{10}[/latex] N/m²
We know that,
Young’s modulus [latex](Y) = \frac{Stress}{Strain}[/latex]
Or, Strain = [latex]Y\times Strain[/latex]
Or, [latex]\frac{T}{A} = Y\times \frac{e}{l} = (9\times 10^{10})\times \frac{5\times 10^{-4}}{1}[/latex]
Or, [latex]\frac{T}{A} = (4.5\times 10^7)[/latex] N/m²
Again
Fundamental frequency [latex](f_1) = \frac{1}{2l}\sqrt{\frac{T}{\mu}}[/latex]
= [latex]\frac{1}{2l}\sqrt{\frac{T}{m/l}}[/latex]

     = [latex]\frac{1}{2l}\sqrt{\frac{Tl}{\rho V}}[/latex]        [[latex]\because m = \rho . V[/latex]]

     = [latex]\frac{1}{2l}\sqrt{\frac{Tl}{\rho Al}} = \frac{1}{2l}\sqrt{\frac{T}{\rho A}}[/latex]

     = [latex]\frac{1}{2l}\sqrt{\frac{Stress}{\rho}}[/latex] [[latex]\because Stress = \frac{T}{A}[/latex]]
= [latex]\frac{1}{2\times 1}\sqrt{\frac{(4.5\times 10^7)}{(9\times 10^3)}}[/latex]
= 35.36 Hz

A wire whose mass per unit length is [latex]10^{-3} kg/m[/latex] is stretched by a load of 4 kg over the two bridge of a sonometer wire 1 m apart. It is struck at its middle point, what would be the wavelength and frequency of its fundamental vibration?

Solution
Here,
mass per unit length of wire, [latex]\mu = 10^{-3} kg/m[/latex]
Load, [latex](m) = 4 kg[/latex]
Distance, [latex]l = 1 m[/latex]
Wavelength, [latex]\lambda = ?[/latex]
Frequency of fundamental vibration, [latex]f = ?[/latex]
We know that,
For fundamental mode,
[latex]\frac{\lambda}{2} = l[/latex]
Or, [latex]\lambda = 2l[/latex]
[latex]\lambda = 2 \times 1 = 2 m[/latex]
Now,
Tension in the wire, [latex]T = mg = 4 \times 10 = 40 N [/latex]
Again,
Fundamental frequency (f)[latex]= \frac{1}{2l}\sqrt{\frac{T}{\mu}}[/latex]
[latex]= \frac{1}{2 \times 1}\sqrt{\frac{40}{10^{-3}}}[/latex]
[latex] = 100 Hz[/latex]

A steel wire of length 20 cm and mass 5 gm is under the tension of 500 N and is tied down at both ends. Calculate the frequency of fundamental mode of vibration.

Solution

Given,

Length of steel wire (L) = 20 cm = 0.2 m

Mass of wire (M) = 5 g = 5 x 10^-3 kg

Tension (T) = 500 N

Frequency of fundamental mode (f) =?

We have,

[latex]\mu = \frac{M}{L}[/latex]

= [latex]\frac{5\times 10^3}{0.2}[/latex]

= 0.025 kg/m

Now,

f = [latex]\frac{1}{2L}\sqrt{\frac{T}{\mu}}[/latex]

= [latex]\frac{1}{2\times 0.2}\sqrt{\frac{500}{0.025}}[/latex]                       

= 353.553 Hz.

A pianofort wire having a diameter of 0.90 mm is replaced by another wire of the same material but with diameter 0.93 mm. If the tension of the wire is as before. What is percentage change in the frequency of fundamental note?

Solution

Given,

Diameter of first wire ([latex]d_1[/latex]) = 0.90 mm = 0.9 x 10^-3 m

Diameter of another wire ([latex]d_2[/latex]) = 0.93 mm = 0.93 x 10^-3 m

Tension is same as before.

% change in the frequency of fundamental note =?

We know,

f = [latex]\frac{1}{2l}\sqrt{\frac{T}{\mu}}[/latex]

            = [latex]\frac{1}{2l}\sqrt{\frac{T}{\frac{\pi d^2}{4}}}[/latex]

            = [latex]\frac{1}{2l}\sqrt{\frac{4T}{\pi d^2}}[/latex]

            = [latex]\frac{(\frac{1}{d_1}-\frac{1}{d_2})}{\frac{1}{d_1}}\times 100%[/latex]

            = [latex]\frac{\frac{d_2 – d_1}{d_1 d_2}}{\frac{1}{d_1}}\times 100%[/latex]

            = [latex]\frac{d_2 – d_1}{d_1 d_2}\times \frac{d_1}{1}\times 100%[/latex]

            = [latex]\frac{d_2 – d_1}{d_2}\times 100%[/latex]

            = [latex]\frac{0.93\times 10^-3 – 0.9 \times 10^-3}{0.93\times 10^-3}\times 100%[/latex]

            = [latex]\frac{0.03}{0.93}\times 100%[/latex]

            = 3.226 %.

On a day when the speed of sound is 345 m/s, the fundamental frequency of a closed organ pipe is 220 Hz. The second overtone of this pipe has the same wavelength as the third harmonic of an open pipe. How long is the open pipe?

Solution

Given,

Speed of sound (v) = 345 m/s

Fundamental frequency of closed organ pipe ([latex]f_1[/latex]) = 220 Hz

Wavelength of second overtone of closed pipe = wavelength of third harmonic of open pipe

Length of open pipe ([latex]l_o[/latex]) =?

Second overtone of closed pipe is 5th harmonic.

i.e. frequency for second overtone ([latex]f_3[/latex]) = [latex]5f_1[/latex]

[latex]= 5 \times 220[/latex]

[latex]= 1100\ Hz[/latex]

So, wavelength of second overtone ([latex]\lambda_3[/latex]) = [latex]\dfrac{v}{f_3}[/latex]

[latex]= \dfrac{345}{1100} = 0.314\ m[/latex]

Now, wavelength of third harmonic of open pipe ([latex]\lambda_3^o[/latex]) =

Or, [latex]\dfrac{3\lambda_3^o}{2} = l_o[/latex]

Or, [latex]l_o = \dfrac{3 \times 0.314}{2} = 0.471\ m[/latex]

= 47.1 cm.

A wire with mass 40 g is stretched so that its ends are tied down at points 80 cm apart. The wire vibrates in its fundamental mode with frequency 60 Hz. Calculate the speed of propagation of transverse waves in the wire and the tension in the wire.

Solution

Given,

Mass of wire (M) = 40g = [latex]40 \times 10^{-3}\ kg[/latex]

Length (L) = 80 cm = 0.8 m

Fundamental frequency (f) = 60 Hz

Speed of transverse wave (v) =?

Tension in the wire (T) =?

We have,

Mass per unit length of wire (μ) = [latex]\dfrac{M}{L}[/latex]

[latex]= \dfrac{40 \times 10^{-3}}{0.8} = 0.05\ kg/m[/latex]

For fundamental mode,

[latex]\dfrac{\lambda}{2} = L[/latex]

Or, [latex]\lambda = 2L[/latex]

[latex]= 2 \times 0.8 = 1.6\ m[/latex]

So, speed (v) = [latex]f \times \lambda[/latex]

[latex]= 60 \times 1.6[/latex]

[latex]= 96\ m/s[/latex]

And,

[latex]f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}[/latex]

Or, [latex]60 = \dfrac{1}{2 \times 0.8}\sqrt{\dfrac{T}{0.05}}[/latex]

Squaring both sides, we get,

[latex]3600 = \dfrac{1}{2.56} \times \dfrac{T}{0.05}[/latex]

Or, [latex]T = 3600 \times 2.56 \times 0.05[/latex]

[latex]= 460.8\ N.[/latex]

A cord of length 1.5 m is fixed at both ends. Its mass per unit length is 1.2 g/m and the tension is 12 N.
(a) What is the frequency of fundamental oscillation?
(b) What tension is required if the n = 3 mode has frequency of 0.50 KHz?

Solution

Given,

Length of cord (l) = 1.5 m

Mass per unit length (μ) = 1.2 g/m = [latex]1.2 \times 10^{-3}\ kg/m[/latex]

Tension (T) = 12 N

(a) Fundamental frequency (f) =?

We know that,

[latex]f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}[/latex]

[latex]= \dfrac{1}{2 \times 1.5}\sqrt{\dfrac{12}{1.2 \times 10^{-3}}}[/latex]

[latex]= 33.333\ Hz[/latex]

(b) T =?, if n = 3 mode has frequency ([latex]f_3[/latex]) = 0.50 KHz = [latex]0.5 \times 10^3\ Hz[/latex]

We know that, for n = 3 mode,

[latex]f_3 = \dfrac{3}{2l}\sqrt{\dfrac{T’}{\mu}}[/latex]

Or, [latex]0.5 \times 10^3 = \dfrac{3}{2 \times 1.5}\sqrt{\dfrac{T’}{1.2 \times 10^{-3}}}[/latex]

Or, [latex]500 = \sqrt{\dfrac{T’}{1.2 \times 10^{-3}}}[/latex]

Squaring both sides,

[latex]250000 = \dfrac{T’}{1.2 \times 10^{-3}}[/latex]

Or, [latex]T’ = 250000 \times 1.2 \times 10^{-3}[/latex]

[latex]= 300\ N.[/latex]

An open pipe 30 cm long and a closed pipe 23 cm long, both of the same diameter, each sounds their first overtone. If they are in resonance, find the end correction of these pipes.

Solution

Given,

Length of open pipe ([latex]l_o[/latex]) = 30 cm

Length of closed pipe ([latex]l_c[/latex]) = 23 cm

End correction (e) =?

For open end pipe, for 1st overtone,

[latex]l_o + 2e = \lambda[/latex] —————— (i)

For closed end pipe, for 1st overtone,

[latex]l_c + e = \dfrac{3\lambda}{4}[/latex]

Or, [latex]\lambda = \dfrac{4(l_c+e)}{3}[/latex] ………………. (ii)

Equating equations (i) and (ii), we get,

[latex]l_o + 2e = \dfrac{4(l_c+e)}{3}[/latex]

Or, [latex]30 + 2e = \dfrac{4(23+e)}{3}[/latex]

Or, [latex]90 + 6e = 92 + 4e[/latex]

Or, [latex]6e – 4e = 2[/latex]

Or, [latex]2e = 2[/latex]

Or, e = 1 cm.

A piano string 1.5 m long is made of steel of density 7800 kg/m³ and Young’s modulus [latex]2 \times 10^{11}\ N/m^2[/latex]. It is maintained at a tension which produced an elastic strain of 1% in the string. Calculate the frequency of transverse vibration of the string when it is vibrating in second mode of vibration.

Solution

Given,

Length of string (l) = 1.5 m

Density (ρ) = 7800 kg/m³

Young’s Modulus (Y) = [latex]2 \times 10^{11}\ N/m^2[/latex]

Strain = 1% = 0.01

Frequency of transverse vibration when vibrating in 2nd mode ([latex]f_2[/latex]) =?

We know that,

[latex]Y = \dfrac{Stress}{Strain}[/latex]

Or, Stress = Y × Strain

Or, [latex]\dfrac{F}{A} = 2 \times 10^{11} \times 0.01[/latex]

Or, [latex]\dfrac{T}{A} = 2 \times 10^9\ N/m^2[/latex] [∵ F = T]

Now, fundamental frequency

[latex]f_1 = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{T}{m/l}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{Tl}{\rho \times V}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{Tl}{\rho \times A \times l}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{T}{\rho A}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\left(\dfrac{T}{A}\right) \times \dfrac{1}{\rho}}[/latex]

[latex]= \dfrac{1}{2 \times 1.5}\sqrt{(2 \times 10^9) \times \dfrac{1}{7800}}[/latex]

[latex]= 168.79\ Hz[/latex]

Again, for 2nd mode ([latex]f_2[/latex]) = [latex]2f_1[/latex]

[latex]= 2 \times 168.79[/latex]

[latex]= 337.58\ Hz.[/latex]

A steel wire of length 40 cm and the diameter 0.25 mm vibrates in unison with a tube, open at both ends and of effective length 60 cm, when each is sounding its fundamental notes. The air temperature is 27°C. Find the tension in the wire. [Given, velocity of sound of air at 0°C = 332 [latex]ms^{-1}[/latex] & density of steel = 7800 [latex]kgm^{-3}[/latex]].

Solution

Given,

Length of steel wire ([latex]l_w[/latex]) = 40 cm = 0.4 m

Diameter ([latex]d_w[/latex]) = 0.25 mm = [latex]0.25 \times 10^{-3}\ m[/latex]

Length of tube ([latex]l_t[/latex]) = 60 cm = 0.6 m

Temperature ([latex]T_1[/latex]) = 27°C = 27 + 273 = 300 K

Tension in the wire (T) =?

Velocity of sound of air ([latex]v_2[/latex]) = 332 m/s

Temperature ([latex]T_2[/latex]) = 0°C = 0 + 273 = 273 K

Density of steel (ρ) = 7800 [latex]kgm^{-3}[/latex]

Area of cross section of wire (A) = [latex]\dfrac{\pi d^2}{4}[/latex]

[latex]= \dfrac{22}{7} \times \dfrac{(0.25 \times 10^{-3})^2}{4}[/latex]

[latex]= 4.911 \times 10^{-8}\ m^2[/latex]

For tube, for fundamental frequency,

[latex]\dfrac{\lambda_t}{2} = l_t[/latex]

Or, [latex]\lambda_t = 2l_t = 2 \times 0.6 = 1.2\ m[/latex]

Now, [latex]v \propto \sqrt{T}[/latex]

Or, [latex]\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}}[/latex]

Or, [latex]\dfrac{v_1}{332} = \sqrt{\dfrac{300}{273}}[/latex]

Or, [latex]v_1 = 348.031\ m/s[/latex]

Again,

Fundamental frequency ([latex]f_t[/latex]) = [latex]\dfrac{v_1}{\lambda_t} = \dfrac{348.031}{1.2} = 290.026\ Hz[/latex]

Since, both sounding its fundamental note,

[latex]f_w = f_t = 290.026\ Hz[/latex]

Now, for wire,

[latex]f_w = \dfrac{1}{2l_w}\sqrt{\dfrac{T}{\mu}}[/latex]

[latex]= \dfrac{1}{2l_w}\sqrt{\dfrac{T}{m/l}}[/latex]

[latex]= \dfrac{1}{2l_w}\sqrt{\dfrac{Tl}{m}}[/latex]

[latex]= \dfrac{1}{2l_w}\sqrt{\dfrac{Tl}{\rho \times V}}[/latex]

[latex]= \dfrac{1}{2l_w}\sqrt{\dfrac{Tl}{\rho \times A \times l}}[/latex]

[latex]= \dfrac{1}{2l_w}\sqrt{\dfrac{T}{\rho \times A}}[/latex]

Or, [latex]290.026 = \dfrac{1}{2 \times 0.4}\sqrt{\dfrac{T}{7800 \times 4.911 \times 10^{-8}}}[/latex]

Or, [latex]232.021 = \sqrt{\dfrac{T}{3.831 \times 10^{-4}}}[/latex]

Squaring both sides,

[latex]5.383 \times 10^4 = \dfrac{T}{3.831 \times 10^{-4}}[/latex]

Or, [latex]T = 20.623\ N.[/latex]

A piano string has length of 2 m and a density of 800 [latex]kgm^{-3}[/latex]. When the tension in the string produces a strain of 1%, the fundamental note obtained from the string in the transverse vibrations in 170 Hz. Calculate Young’s modulus for the material of the string.

Solution

Given,

Length of string (l) = 2 m

Density (ρ) = 800 kg/m³

Strain = 1% = 0.01

Young’s Modulus (Y) =?

Frequency of fundamental vibration (f) = 170 Hz

We know that,

[latex]Y = \dfrac{Stress}{Strain}[/latex]

Or, Stress = Y × Strain

Or, [latex]\dfrac{F}{A} = Y \times 0.01[/latex]

Or, [latex]\dfrac{T}{A} = 0.01Y[/latex] [∵ F = T]

Now, fundamental frequency

latex = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{T}{m/l}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{Tl}{\rho \times V}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{Tl}{\rho \times A \times l}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{T}{\rho A}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\left(\dfrac{T}{A}\right) \times \dfrac{1}{\rho}}[/latex]

Or, [latex]170 = \dfrac{1}{2 \times 2}\sqrt{0.01Y \times \dfrac{1}{800}}[/latex]

Or, [latex]680 = \sqrt{\dfrac{0.01Y}{800}}[/latex]

Squaring both sides,

[latex]4.624 \times 10^5 = \dfrac{0.01Y}{800}[/latex]

Or, [latex]0.01Y = 3.699 \times 10^8[/latex]

Or, [latex]Y = \dfrac{3.699 \times 10^8}{0.01} = 3.699 \times 10^{10}\ Nm^{-2}.[/latex]

A wire of diameter 0.04 cm and made of steel of density 8000 [latex]kgm^{-3}[/latex] is under constant tension of 80 N. What length of this wire should be plucked to cause it to vibrate with a frequency of 840 Hz?

Solution

Given,

Diameter of wire (d) = 0.04 cm = [latex]0.04 \times 10^{-2}\ m[/latex]

Density (ρ) = 8000 [latex]kgm^{-3}[/latex]

Tension (T) = 80 N

Length of wire (l) =?

Frequency (f) = 840 Hz

We know that,

[latex]f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{T}{m/l}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{Tl}{\rho \times V}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{Tl}{\rho \times A \times l}}[/latex]

[latex]= \dfrac{1}{2l}\sqrt{\dfrac{4T}{\rho \times \pi d^2}}[/latex]

Or, [latex]840 = \dfrac{1}{2l}\sqrt{\dfrac{4 \times 80}{8000 \times 3.14 \times (0.04 \times 10^{-2})^2}}[/latex]

Or, [latex]840 = \dfrac{2.822 \times 10^2}{2l}[/latex]

Or, [latex]1.68 \times 10^3 l = 2.822 \times 10^2[/latex]

Or, [latex]l = \dfrac{2.822 \times 10^2}{1.68 \times 10^3} = 0.168\ m.[/latex]

An organ pipe is tuned to a frequency of 440 Hz when the temperature is 27°C. Find its frequency when the temperature drops to 0°C. Assume both ends of the pipe open.

Solution

Given,

Frequency ([latex]f_1[/latex]) = 440 Hz

Temperature ([latex]T_1[/latex]) = 27°C = 27 + 273 = 300 K

Frequency ([latex]f_2[/latex]) =?

Temperature ([latex]T_2[/latex]) = 0°C = 0 + 273 = 273 K

We know that,

[latex]v \propto \sqrt{T}[/latex]

Or, [latex]\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}}[/latex]

Or, [latex]\dfrac{f_1 \times \lambda}{f_2 \times \lambda} = \sqrt{\dfrac{T_1}{T_2}}[/latex]

Or, [latex]\dfrac{f_1}{f_2} = \sqrt{\dfrac{T_1}{T_2}}[/latex]

Or, [latex]\dfrac{440}{f_2} = \sqrt{\dfrac{300}{273}}[/latex]

Or, [latex]440 = 1.048f_2[/latex]

Or, [latex]f_2 = 419.847\ Hz.[/latex]

A steel wire of 2 m long whose mass is 3g is under tension of 500 N and is tied down at both ends. Calculate the frequency and wavelength for fundamental mode of vibration.

Solution

Given,

Length of wire (l) = 2 m

Mass of wire (m) = 3g = [latex]3 \times 10^{-3}\ kg[/latex]

Tension (T) = 500 N

Frequency of fundamental vibration (f) =?

Wavelength (λ) =?

We know that,

Mass per unit length (μ) = [latex]\dfrac{m}{l}[/latex]

[latex]= \dfrac{3 \times 10^{-3}}{2} = 1.5 \times 10^{-3}\ kg/m[/latex]

Now, fundamental frequency (f) = [latex]\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}[/latex]

[latex]= \dfrac{1}{2 \times 2}\sqrt{\dfrac{500}{1.5 \times 10^{-3}}}[/latex]

[latex]= 144.338\ Hz[/latex]

Again, for fundamental mode,

[latex]\dfrac{\lambda}{2} = l[/latex]

Or, [latex]\lambda = 2l[/latex]

[latex]= 2 \times 2[/latex]

[latex]= 4m.[/latex]