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Hatshy [7]
3 years ago
15

The tension of a guitar string is increased by 40%. By what factor odes the fundamental frequency of vibration change? a. 1.13 b

. 1.18 c. 1.06 d. 1.09 e. 1.24
Physics
1 answer:
bogdanovich [222]3 years ago
8 0

Answer:

<h3> b. 1.18</h3>

Explanation:

The fundamental frequency in string is expressed as;

F1 = 1/2L√T/m .... 1

L is the length of the string

T is the tension

m is the mass per unit length

If the tension is increased by 40%, the new tension will be;

T2 = T + 40%T

T2 = T + 0.4T

T2 = 1.4T

The new fundamental frequency will be;

F2 = 1/2L√1.4T/m ..... 2

Divide 1 by 2;

F2/F = (1/2L√1.4T/m)/1/2L√T/m)+

F2/F = √1.4T/m ÷ √T/m

F2/F = √1.4T/√m ×√m/√T

F2/F = √1.4T/√T

F2/F = 1.18√T/√T

F2/F = 1.18

F2 = 1.18F

Hence the fundamental frequency of vibration changes by a factor of 1.18

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Answer with Explanation:

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Using the formula

4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}

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n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}

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n=\frac{N}{L}

L=\frac{N}{n}

L=\frac{799}{6404}

L=0.125 m=0.125\times 100=12.5 cm

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