94% Sun Apr 10 10 54 38 PM a E Chrome File Edit View History Bookmarks People Wi

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94% Sun Apr 10 10 54 38 PM a E Chrome File Edit View History Bookmarks People Window Help A Josh Chapter 17 Linear S C Books C My Internships I Cheg C Physics question I Ch google maps Google c www.webassign.net 13047796 /web/Student/Assignment-Responses/last?dep App Bookmarks Apple YouTube home Watch Live Sports E O tech Destiny LFG Net l The faculty.pnc.edu/ptes Screen Shot PM 2016-04...0.08 PM Screen Shot PM 2016-04...0.20 PM 5, C+ -l2 points CJ10 17.P.053 My Notes Ask Your Teacher A string is fixed at both ends and is vibrating at 144 Hz, which is its third harmonic frequency. The linear density of the string is 5.45 x 10 3 kg/m, and it is under a tension of 3.1 N. Determine the length of the string. o your Screen Shot Additional Materials 2 PM 2016-04...0.26 PM ct you a Section 17 Screen Shot PM 2016-04...0.30 PM 6. C+ -l2 points CJ10 17. 027 My Notes Ask Your Teacher his The fundamental frequency of a string fixed at both ends is 260 Hz. How long does it take for a wave to travel the length of this s ring? Screen Shot PM 2016-04...0.40 PM Additional Materials ection 17.5 Screen Shot PM 2016-03...4.06 PM 7. -12 points CJ10 17.P.029 My Notes Ask Your Teacher

Explanation / Answer

5)

If T is the tension, M is the linear density, and v is the speed, then

v = sqrt(T/M)

So the speed of the wave = sqrt(3.1/(5.45*10^(-3)) = 23.85 m/s.

Now with any wave,

speed = frequency times wavelength

So wavelength = v/f = (23.85)/144 = 0.1656 is the wavelength of the wave.

Now, we are at the third harmonic frequency. The string is fixed at both ends, so therefore it must be that the string is an exact multiple of half a wavelength long.

The first harmonic is when the string is exactly half a wavelength long. The second one is when the string is exactly 1 full wavelength long. The third one is when the string is exactly 1.5 wavelengths long,

So the string must be 1.5*0.1656 = 0.2484 m long

6)

fundamental frequency means we only have half of one period of a wave

1/frequency = period

1/260 = time it takes to complete a whole wave

half of this or 1/520 seconds is the time it takes to travel the length of the string
approximately 1.92ms

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