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A string, fixed at both ends, has a length of 6 m and supports a standing wave with a total of 4 nodes. If a transverse wave can travel at 40 m/s down the rope, what is the frequency of this standing wave?(A) 6.7 $\mathrm{Hz}$(B) 10 $\mathrm{Hz}$(C) 20 $\mathrm{Hz}$(D) 26.7 $\mathrm{Hz}$

B

Physics 101 Mechanics

Physics 102 Electricity and Magnetism

Chapter 9

Waves

Sound and Hearing

Mechanical Waves

Electromagnetic Waves

University of Washington

Simon Fraser University

University of Sheffield

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it's problem. We're gonna talk about the frequency and wavelength off waves. Eso consider that we have string like this one shown in the picture we call nodes. These points of minimum amplitude here that I'm marking in rep. Okay, so if this is a standing wave, these points we'll always have zero amplitude and the distance between, uh, two nonconsecutive consecutive points, as shown here in the picture is equal to the wave one. Okay, so this year is one wavelength off the week. Yeah. Okay. Also, we know that the frequency times the wavelength is equal to the speed of replication of the waves. V In our problem, we have ah, string that has a length off 6 m and has four nodes. So the string would look something like this. Okay, so notice that there are four notes here. 1234 uh, in the length is equal to 6 m. And the exercise tells us that the speed of propagation of the waves is it which is 40 m per second. And our goal is to find what is the frequency of the weight. Now, notice that in this wave we have 1.5 wavelength because this year is one up wavelength. This here is half a wavelength. Eso lum the f I'm sorry. So one 0.5 times the wavelength is able to 6 m. So long Day six divided by 1.5. That's 4 m. This is the This is the wavelength on the weight now F times Lunda is V so f is London video Very, very London. This means that F is equal to 40 m per second, divided by Lambda, which is 4 m. So AF is equal to 10 Hertz, which is our answer, disagrees with option B in the alternatives.

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