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$\bullet$ When a mass $M$ hangs from a vertical wire of length $L$waves travel on this wire with a speed $V .$ What will be thespeed of these waves (in terms of $V$ ) if (a) we double $M$ without stretching the wire? (b) we replace the wire with an identical one, except twice as long? (c) we replace the wire withone of the same length, but three times as heavy? (d) westretch the wire to twice its original length? (e) we increase$M$ by a factor of $10,$ which stretches the wire to double itsoriginal length?

a) $\sqrt{2} V$b) $v_{2}=v$c) $\sqrt{\frac{1}{3}} \times v$d) $\sqrt{2} v$e) 2$\sqrt{5} v$

Physics 101 Mechanics

Chapter 12

Mechanical Waves and Sound

Periodic Motion

Mechanical Waves

Sound and Hearing

Simon Fraser University

University of Winnipeg

McMaster University

Lectures

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Okay, so first, let's take a look at the equation here, so we know a p v committee of a mule tease, the tention muse, the mass per union length and the mass per unit length can be expressed as move, which is the mass of wire over the Length of wire, so if you do some arrangement here, we'll have squareroot over m and in this case tension can we go to capital m gwish is the gravity of the mass m so eventually have sped v here is equal to square root capital m g over Lower case m so for the first case, so we double the mass. So therefore we have a new speed here, which is v. Prime is equal to square root, 2 capital m g l over lower case, and this will give us 2 times capital m g l over m p, as you can tell this turn here, is equal to the initial speed which is here so therefore, prime, is equal To 2, so the new speed value should be a to square root 2, and for the next case, so you were saying that it was saying that the length of y was double but its identical 1, which means that the mass per union length is the same. So if the length is double, which means that the mass of the wire should be double as well, okay, because the mass for union is the same so therefore we have a new speed here, which is v. Prime is equal to square root capital m g times 2. L over 2 lower case m, as you can tell to her, can be cancel out so eventually we'll just have square root. Capital m g l over the case m, which is equal to v. So, therefore, the new speed is equal to the initial speed and for the next case, so in this case the mass of wire was triple so therefore he he new speed. We prime here is equal to square root. Capital m g l over the new mass of the wire, which is 3 m, and this can be good to square root. 1 over 3 times square root. Capital m g l over m, as you can tell this term here- is equal to initial p v. So therefore, we have v prime, is equal to square root 1 over 3 times capital v here, and for this case so for this case the length of the wire was double so therefore, v prime, is equal to square root. Capital m g times 2 l over lower case m, which is equal to squat times square root. Capital m g, l over lower case m, and this will give us the new speed- is equal to square to okay. Because of this turn here is equal to v. So, for the last case, so now the mass m is increased by a factor over tan and the lens is double so lef have a new speed. We prime here is equal to 10 capital m g l, o g times 2 l, okay, becase the lent with doane over look case m, as you can tell it's on, be good to square root: 20 times square root; capital m g, l over low case m Square root 20 year is equal to 2 square root 5 and square capital m g over lower case m here is equal to v. So therefore have the new speed v prime, is equal to 25 times capital v, and these are the answers for this question.

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