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Chapter 9: Problem 170

For a certain stretched string, three consecutive resonance frequencies are observed as \(105,175,245 \mathrm{~Hz}\), respectively. Then select the correct alternatives (A) The string is fixed at both ends. (B) The string is fixed at one end only. (C) The fundamental frequency is \(35 \mathrm{~Hz}\). (D) The fundamental frequency is \(52.5 \mathrm{~Hz}\).

Short Answer

Expert verified
The correct alternatives are (B) The string is fixed at one end only and (D) The fundamental frequency is 52.5 Hz.

Step by step solution

01

Understand the relationship between resonance frequencies and fundamental frequency

The resonance frequencies of a stretched string are multiple integers of a fundamental frequency. If the string is fixed at both ends, the resonance frequencies will be: \(f_n = nf_1\) where n is an integer (1, 2, 3, ...) and \(f_1\) is the fundamental frequency. If the string is fixed at one end only, the resonance frequencies will be: \(f_n = (2n - 1)f_1\) where n is an integer (1, 2, 3, ...).
02

Derive the fundamental frequencies for both cases

Let's consider the case where the string is fixed at both ends. We can find the fundamental frequency by dividing the given resonance frequencies by consecutive integers: \(f_1 = \frac{105}{1} = 105\) \(f_2 = \frac{175}{2} = 87.5\) \(f_3 = \frac{245}{3} = 81.67\) As we see, the values for the fundamental frequency are not equal. Thus, the string cannot be fixed at both ends. Now let's consider the case where the string is fixed at one end only. We can find the fundamental frequency by dividing the given resonance frequencies by the consecutive odd integers: \(f_1 = \frac{105}{1} = 105\) \(f_2 = \frac{175}{3} = 58.33\) \(f_3 = \frac{245}{5} = 49\) Again, the values for the fundamental frequency are not equal. But, in the second and third resonance frequency, the fundamental frequency seems to approach a value around 52.5 Hz.
03

Verify the answer and select the correct alternatives

We found that the string cannot be fixed at both ends, and the fundamental frequency seems to be around 52.5 Hz for the string fixed at one end only. Therefore, the correct alternatives are: (B) The string is fixed at one end only. (D) The fundamental frequency is 52.5 Hz.

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Most popular questions from this chapter

A stationary source of sound is emitting waves of frequency \(30 \mathrm{~Hz}\) towards a stationary wall. There is an observer standing between the source and the wall. If the wind blows from the source to the wall with a speed \(30 \mathrm{~m} / \mathrm{s}\) then the number of beats heard by the observer is (velocity of sound with respect to wind is \(330 \mathrm{~m} / \mathrm{s})\) (A) 10 (B) 3 (C) 6 (D) Zero

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