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A string of length L, mass per unit length \mu, and tension T is vibrating at its fundamental frequency. What effect will the following have on the fundamental frequency?

(a) The length of the string is doubled, with all other factors held constant ?

(b) The mass per unit length is doubled, with all other factors held constant.

(c) The tension is doubled, with all other factors held constant.

User Shcheklein
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2 Answers

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Final answer:

The length of the string directly affects the fundamental frequency, while the mass per unit length and tension have different effects on the fundamental frequency.

Step-by-step explanation:

The fundamental frequency of a vibrating string is determined by its length, mass per unit length, and tension. Let's analyze the effects of the given changes:

(a) When the length of the string is doubled while keeping other factors constant, the fundamental frequency will be halved. This is because the wavelength of the fundamental mode is directly proportional to the length of the string.

(b) When the mass per unit length of the string is doubled while keeping other factors constant, the fundamental frequency will remain unchanged. The mass per unit length does not affect the frequency of the fundamental mode.

(c) When the tension of the string is doubled while keeping other factors constant, the fundamental frequency will increase. This is because the tension is inversely proportional to the wavelength of the fundamental mode.

User Dhana
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The fundamental frequency on a vibrating string is given by:


f=(1)/(2L)\sqrt{(T)/(\mu)}

where

L is the length of the string

T is the tension


\mu is the mass per unit length of the string

Keeping this equation in mind, we can now answer the various parts of the question:

(a) The fundamental frequency will halve

In this case, the length of the string is doubled:

L' = 2L

Substituting into the expression of the fundamental frequency, we find the new frequency:


f'=(1)/(2(2L))\sqrt{(T)/(\mu)}=(1)/(2)((1)/(2L)\sqrt{(T)/(\mu)})=(f)/(2)

So, the fundamental frequency will halve.

(b) the fundamental frequency will decrease by a factor
√(2)

In this case, the mass per unit length is doubled:


\mu'=2\mu

Substituting into the expression of the fundamental frequency, we find the new frequency:


f'=(1)/(2L)\sqrt{(T)/(2 \mu)}=(1)/(√(2))((1)/(2L)\sqrt{(T)/(\mu)})=(f)/(√(2))

So, the fundamental frequency will decrease by a factor
√(2).

(c) the fundamental frequency will increase by a factor
√(2)

In this case, the tension is doubled:


T'=2T

Substituting into the expression of the fundamental frequency, we find the new frequency:


f'=(1)/(2L)\sqrt{(2T)/(\mu)}=√(2)((1)/(2L)\sqrt{(T)/(\mu)})=√(2)f

So, the fundamental frequency will increase by a factor
√(2).

User Maxim Norin
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