Numerical 1 [2082]
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. [3]
i. stress developed in the string.
ii. frequency of second mode of vibration.
Ans: [latex]2\times 10^9[/latex] N/m2, 337 Hz
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
Numerical 2 [2082 GIE ‘A’]
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]. [2] Ans: 35.36 Hz
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
Numerical 3 [2073 ‘B’]
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]
Numerical 4 [2074 ‘B’]
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.
Numerical 5 [2073 ‘S’]
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 %.