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12345 [234]
1 year ago
11

A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 hz. what driving freq

uency will set up a standing wave with five equal segments?
Physics
1 answer:
Korvikt [17]1 year ago
8 0

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.

<h3>How to find the driving frequency?</h3>
  • 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

#SPJ4

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If a grocery cart with a mass of 16.5 kg accelerates at +2.31 m/s2 against a frictional force of -15.0 N, what is the applied fo
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Call the applied force 'A'. (Clever ?)

The forces on the cart are  'A' forward and 15 N backward.

The net force on the art is (A-15) forward.

F = m a

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(A - 15) = (16.5) x (2.31)

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