• Support PF! Buy your school textbooks, materials and every day products Here!

Understanding the equation of an oscillating string

  • #1

Homework Statement



y' = (0.80 cm) sin[(π/3 cm-1)x] cos[(45π s-1)t]
(a) What are the amplitude and speed of the two waves (identical except for direction of travel) whose superposition gives this oscillation?
(b) What is the distance between nodes?
(c) What is the speed of a particle of the string at the position x = 1.5 cm when t = 9/8 s?

Homework Equations


(a)y1=(.8cm)sin[(π/3 cm-1)x + (45π s-1)t]
y2=(.8cm)sin[(π/3 cm-1)x - (45π s-1)t]

y(x,t)=A*sin(kx-omega*t)
where A is ampltude, k is wave number, omega is frequency

(b)lambda=2π/k

The Attempt at a Solution


When I answered that my amplitude was .8cm, that was wrong. so I'm confused about whether I'm using the wrong interpretation of the oscillating string's equation. Also, when I used the equation in (b) I got the wrong wavelength. I think the problem is that I'm not understanding what each of the numbers in the original equation correspond to.
 

Answers and Replies

  • #2
ehild
Homework Helper
15,478
1,854
This is a standing wave, the sum of two sine waves travelling in opposite directions. Find these two waves.


ehild
 
  • #3
I thought that was the first thing I did under useful equations. Are those two equations not right for the two waves traveling in opposite directions?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955

Homework Statement



y' = (0.80 cm) sin[(π/3 cm-1)x] cos[(45π s-1)t]
(a) What are the amplitude and speed of the two waves (identical except for direction of travel) whose superposition gives this oscillation?
(b) What is the distance between nodes?
(c) What is the speed of a particle of the string at the position x = 1.5 cm when t = 9/8 s?

Homework Equations


(a)y1=(.8cm)sin[(π/3 cm-1)x + (45π s-1)t]
y2=(.8cm)sin[(π/3 cm-1)x - (45π s-1)t]
The amplitudes are NOT the same as the amplitude of the original function, .8.
sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y)
sin(x- y)= sin(x)cos(y)- cos(x)sin(y) (because sine is an odd function and cosine even)

Adding the two equations, 2 sin(x)cos(y)= sin(x+ y)+ sin(x- y) and so
Asin(x)cos(y)= (A/2)sin(x+y)+ (A/2)sin(x- y).

y(x,t)=A*sin(kx-omega*t)
where A is ampltude, k is wave number, omega is frequency

(b)lambda=2π/k

The Attempt at a Solution


When I answered that my amplitude was .8cm, that was wrong. so I'm confused about whether I'm using the wrong interpretation of the oscillating string's equation. Also, when I used the equation in (b) I got the wrong wavelength. I think the problem is that I'm not understanding what each of the numbers in the original equation correspond to.
 

Related Threads on Understanding the equation of an oscillating string

Replies
3
Views
554
Replies
9
Views
163
Replies
10
Views
11K
Replies
30
Views
903
Replies
1
Views
4K
Replies
0
Views
2K
  • Last Post
Replies
15
Views
1K
  • Last Post
Replies
2
Views
774
  • Last Post
Replies
9
Views
13K
Replies
1
Views
2K
Top