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A certain guitar string has a mass of 2.0 g and a length of 60 cm. What must the tension in the string be if the speed of the wave on it is to be 300 m/s?...

Question

A certain guitar string has a mass of 2.0 g and a length of 60 cm. What must the tension in the string be if the speed of the wave on it is to be 300 m/s?

A certain guitar string has a mass of 2.0 g and a length of 60 cm. What must the tension in the string be if the speed of the wave on it is to be 300 m/s?



Answers

A string has a mass of $3.0 \mathrm{~g}$ and a length of $60 \mathrm{~cm}$. What must be the tension so that when vibrating transversely its first overtone has frequency $200 \mathrm{~Hz}$ ?

Students in this problem, we have to find what is the length of that string? We have to find the length. So here we have given the mass of the string is five into 10 to the power minus three kg. And the tension Yeah. Is 1 80 Newton? Yeah. And frequency of wave, which is traveling, is to 60. Hunch. Yeah, and violence is also given, which is 0.60 m. So first of all, we will try to find the speed of the way we know that Formula Spear is equal to f into Lambda. Thank you. Night We are f is the frequency and lambda is the violence eight and mhm. Mhm. This is one formula and another formula of speed as a quality Underwood, Steve, I knew were news that linear mass density which is am by and yeah, so we will use both this formula. So, using this formula, we can write f lambda should be called to under road TV mu which you is a call to em by l. So from here l Should we call to f Square Lamba? Square em by P. So frequency is 2 60. Lambda is 0.60 and Mars is five into the end of the power minus three. And tension is 1 18 years. So the length of the vibe should be. If we solve this. This is a cold around 0.676 meter. Yeah.

For this problem on the topic. Off waves and sound were given the mass off a string as well as its tension. When it is stretched, we're told that it transfers, wave them, traveled on the string with a given frequency and wavelength and were asked to calculate the length of the string. Now we know the length of the string is one factor that effects the speed, awful way of traveling on it. And the equation relating the two V is equal to the square. Root off the attention on the stream f divided by the mass per unit length off the string em off all so we consider the other factor affecting the speed is detention F. Now we're not given the speed in this problem we But we are given the frequency and wavelength. And we know that the is also equal to the frequency times in Waveland so we can equate need to these two expressions. So the speed of the wave is times lambda and also the square root off f over m times l, which means if we rearrange this equation, we can solve for the length off the string as required, so really engine we get the lent out is able to be frequency squared times three Waveland squared times the mass of the string divided by detention and all of these values unknown. So we can find this length. That's 260 hertz frequency squared and the wavelength off the wave. 0.6 m and that squared times the mass of the string which is five times 10 to the minus three kg. All over the tension in the string which is given to be 180 Newtons and so calculating, we get the length off the string to be 0.68 meters.

How to find the velocity of the wave on a string that has tension Teoh and then your mass density. Well, let's say that tension is 90 mins and is 3.2 grounds. The first thing you have to do is how to stop. This is N Grams. So you have to convert two kilograms because Newton's is killed grounds times meter, over second square. So if you convert this two kilograms, you have to follow. It's going to be devoured by thousands. No, this this no fuel grounds over meters. Not so it use talk this 19 even in the final zero. Does your fear 32 Yeah. Remember these all scurvy and no free units, no grounds who comes for me. So when you do this 90 divided by your 900.32 we had about 20 it down so you could take the square root of that. You'll end up with 167 meters per second

Yeah. This question where the concept of the web speed in the string and the speed is given a spiral out of the tension in the string upon the linear mass density of the string. So it's for part a The spirit is equivalent spiraled out. The tension in the tension is 62 Point to Newton upon to linear mass density. That is 1.93. You two vendors -3 kilograms for the web speed is curing too. Uh 180 m for second. No part B. So from the equation we have used in the concept, we can say the force is a covering two. We square into the reading must entertain you. And in this situation the speed we even is the Cuban. 1.01. Thanks. Do you tell us food? We so if for we even the force of attention afghani security, we won square into milk. Or we can write, this is 1.01 square into the square into New. I reckon. Right. Have fun as uh 1.0 201 times initial force F, Which means the changing force is the equivalent to 2.01%. So to increase the speed by 1%, the attention in the stream must be changed by 2.0.


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