Question

What is the maximum transverse speed v_y of the bead? For a standing wave.

Answer 1

I have done everything and am on the last final step but I am still unable to solve the problem.

Answer 2

@ProfBrainstorm, I have followed the steps but am still having trouble.

Answer 3

You know the displacement of the string is Asin(wt)sin(kx)

Have you figured out correctly the expression for the transverse velocity ?

Answer 4

Awsin(kx)cos(wt)

Answer 5

(7.50*10^-2)(5.71*10.5)sin(5.71x)cos(5.71*10.5*t) am I not writing it out properly?

Answer 6

Yes, now, you have the value of k, do you also have the value of omega ?

Answer 7

w=vk
w=(10.5*5.71)
w=59.95

Answer 8

Alright, now we have everything we need except the value of x to plug in, what value are you using for x ?

Answer 9

In other words, what is the x coordinate of the bead along the string ?

Answer 10

sorry I was making waffles!

Answer 11

I wasn't sure, but should x even matter if it's always going to be between 1 and -1

Answer 12

x very much matters, it is a fixed value at the position of the bead, it is the other term involving the time that we can essentially ignore

Answer 13

x is not the transverse displacement of the bead, it is simply the distance along the string

Answer 14

27.5!

Answer 15

silly me

Answer 16

no

Answer 17

ouch

Answer 18

the first antinode occurs at x=0.275 m

Answer 19

and then the bead is positioned 0.138 m to the right of that antinode, so . . .

Answer 20

conversion and subtraction! thanskingiving has hit me hard.

Answer 21

waffles, thanksgiving : (

Answer 22

in my country. but .275-.138=.137 :^)?

Answer 23

i think you need to add ?

Answer 24

(7.50*10^-2)(5.71*10.5)sin(5.71*0.413)cos(5.71*10.5*t)

Answer 25

that first factor should be7.5 * 10^-3, should it not ? converting from mm to m ?

Answer 26

yes yes that is correct o 0.0075

Answer 27

5.71*10.5 is w

Answer 28

okay, now do you understand that the cosine involving the time can just be set equal to 1, in order to find the maximum value of the expression as time varies ?

Answer 29

so 1 to find the max value as time varies, yes.

Answer 30

alright, so now just calculate what the rest of it comes to and you have v max

Answer 31

http://www.wolframalpha.com/input/?i=%287.50*10%5E-3%29%285.71*10.5%29sin%285.71*0.413%29cos%285.71*10.5*1%29

Answer 32

No, you've misunderstood
You don't just set t=1, you set the whole cosine expression involving time equal to 1

Answer 33

"Since all of the other terms are constant over time, at a given location, the maximum value for vy(x,t) will occur when this time-dependent sine term is at a maximum. Thus, you simply need to know that the sine function oscillates between ±1." This is what they are telling me as a hint

Answer 34

Yes, I understand what they are saying, that's what I'm telling you as well
You need to work out (7.50*10^-3)(5.71*10.5)sin(5.71*0.413)

Answer 35

ohhhhhh light bulb, I've answered my own questio.

Answer 36

im so silly. No I get it now :^)

Answer 37

Okay, but one last point where you have to be careful, the quantity in the remaining sine is not in degrees, it is in radians, so be careful with units when evaluating the sine

Answer 38

Let me know what final answer you get

Answer 39

0.317312 rounded to 0.314

Answer 40

i mean 0.317

Answer 41

0.317 is what I have too, yes : ) meters per second, don't forget the units !

Answer 42

Bingo bingo! Correct

Answer 43

glad i could help

Answer 44

Part D
To check your equation for the standing wave's transverse velocity, find the maximum transverse velocity at x = 55.0cm .
Express your answer in meters per sec

Answer 45

Don't help jsut yet, I am doing

Answer 46

ok, i'll check back later

Answer 47

Thank you for your help!

Answer 48

so transverse velocity at x=55cm, i plug in 0.55 into x into the previous equation and get a very small number: 0.000491325

Answer 49

hmmm - interesting
what was the position of that first antinode again ?

Answer 50

27.5cm

Answer 51

woah!

Answer 52

right, and if you move out twice as far, you come to a node, a point where there is no motion whatsoever in the standing wave

Answer 53

you see what they're getting at ?

Answer 54

So I added and it's magic, I got -0.318 this time

Answer 55

well if you move out to x=55cm, you've arrived at a node, so there should be no motion, zero velocity, which is what I found

Answer 56

when you say you added, what do you mean ?

Answer 57

or 0.000491325? I got -0.318 adding .138 to .55 with the light bead.

Answer 58

or is it just .55? if so then it is 0.000491325

Answer 59

ah no, it's nothing to do with the bead now, they're telling you right off that x=o.55m

Answer 60

so 0 or 0.000491325

Answer 61

well zero to the accuracy that we've been using

Answer 62

Yes that makes sense!

Answer 63

you see what they're getting you to do - you know that there is a node at x=55cm and so they ask you to see if the equation predicts this zero motion position correctly

Answer 64

which it does

Answer 65

yes no motion being there!

Answer 66

right

Answer 67

eureeka, nail on the head. Thank you again!

Answer 68

welcome