Question

A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 hz. what driving frequency will set up a standing wave with five equal segments?

Answer 1

600Hz is the driving frequency needed to create a standing wave with five equal segments.

To find the answer, we have to know about the fundamental frequency.

How to find the driving frequency?

• The following expression can be used to relate the fundamental frequency to the driving frequency;

                                        f(n) = n * f (1)

where, f(1) denotes the fundamental frequency and the driving frequency f(n).

• The standing wave has four equal segments, hence with n=4 and f(n)=4, we may calculate the fundamental frequency.

                                          f(4) = 4× f (1)

                                          480 = 4× f(1)

                                         f(1) = 480/4 =120Hz.

So, 120Hz is the fundamental frequency.

• To determine the driving frequency necessary to create a standing wave with five equally spaced peaks?
• For, n = 5,

                      f(n) = n 120Hz,

                      f(5) = 5×120Hz=600Hz.

Consequently, 600Hz is the driving frequency needed to create a standing wave with five equal segments.

Learn more about the fundamental frequency here:

brainly.com/question/2288944

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