Understanding the equation of an oscillating string

[cosmogirl90]

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.

 

[ehild]

This is a standing wave, the sum of two sine waves traveling in opposite directions. Find these two waves. ehild

 

[cosmogirl90]

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?

 

[HallsofIvy]

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.

 

[rwisz]

I can provide some clarification on the equation of an oscillating string. The equation you provided is a general form of the wave equation, where y' represents the displacement of the string at position x and time t. The terms within the sine and cosine functions represent the wave number (k), frequency (ω), amplitude (A), and phase shift (φ).

To answer the questions provided, we need to first determine the values of these terms. The amplitude can be found by taking the maximum value of the displacement, which in this case is 0.80 cm. The speed of the wave can be calculated by dividing the frequency (45π s-1) by the wave number (π/3 cm-1), resulting in a speed of 135 cm/s.

To find the distance between nodes, we can use the equation λ=2π/k, where λ represents the wavelength. Plugging in the values for k and solving for λ, we get a wavelength of approximately 6.28 cm. This means that the distance between consecutive nodes is 6.28 cm.

To determine the speed of a particle at a specific position and time, we can use the general equation y(x,t)=A*sin(kx-ωt). Plugging in the values for x=1.5 cm and t=9/8 s, we get a displacement of approximately 0.57 cm. This means that the particle at x=1.5 cm has a speed of 0.57 cm/9/8 s, which is approximately 0.64 cm/s.

It's important to note that the values of amplitude, frequency, and wave number can change depending on the specific system and conditions being studied. It's always important to carefully analyze and understand the equation being used in order to accurately interpret the results.