- #1
Bassa
- 46
- 1
λ∂
Aguitar string lies along the x-axis when in equilibrium. The end of
the string at x=0 (the bridge of the guitar) is fixed. A sinusoidal
wave with amplitude A=0.750 mm and frequency
f =440 Hz, corresponding to the red curves in Fig. 15.24,
travels along the string in the -x-direction at 143m/s. It is reflected
from the fixed end, and the superposition of the incident and reflected
waves forms a standing wave. (a) Find the equation giving the displacement
of a point on the string as a function of position and time.
(b) Locate the nodes. (c) Find the amplitude of the standing wave and
the maximum transverse velocity and acceleration.
∂y(x,t)/∂t=AswSin(kx)Cos(wt)w
T=1/f
λ=v/f
position of a node=λ/2
So, I get how to do everything up until part c.[/B]
The partial derivative of the transverse wave with respect to time and holding x constant is:
∂y(x,t)/∂t=(4.15/m/s)sin[(19.3 rad/m)x]cos[(2760rad/sec)t
Now, by just looking at this function, I could tell that the maximum velocity is 4.15. The function will oscillate between +4.15 and -4.15.
Well, I thought that if I find the position of a node and the time at which it will occur, this function would yield to an answer of 4.15m/s.
The values I used for the period is .002sec so, a node will happen at half a period which is .001sec.
One wavelength is .325m and a node will happen half way through. This means that a node will occur at x=.1625m.
When I plug in the values in the above equation I get -.020599061m/s which is not the correct an answer. Would you please let me know if I am thinking about this incorrectly? Doesn't maximum velocity happen at the intersection point with the x-axis? Isn't this point a node in this case? This is an example in a textbook and I am trying to figure the maximum velocity by not just looking at the amplitude of the function.
Homework Statement
Aguitar string lies along the x-axis when in equilibrium. The end of
the string at x=0 (the bridge of the guitar) is fixed. A sinusoidal
wave with amplitude A=0.750 mm and frequency
f =440 Hz, corresponding to the red curves in Fig. 15.24,
travels along the string in the -x-direction at 143m/s. It is reflected
from the fixed end, and the superposition of the incident and reflected
waves forms a standing wave. (a) Find the equation giving the displacement
of a point on the string as a function of position and time.
(b) Locate the nodes. (c) Find the amplitude of the standing wave and
the maximum transverse velocity and acceleration.
Homework Equations
∂y(x,t)/∂t=AswSin(kx)Cos(wt)w
T=1/f
λ=v/f
position of a node=λ/2
The Attempt at a Solution
So, I get how to do everything up until part c.[/B]
The partial derivative of the transverse wave with respect to time and holding x constant is:
∂y(x,t)/∂t=(4.15/m/s)sin[(19.3 rad/m)x]cos[(2760rad/sec)t
Now, by just looking at this function, I could tell that the maximum velocity is 4.15. The function will oscillate between +4.15 and -4.15.
Well, I thought that if I find the position of a node and the time at which it will occur, this function would yield to an answer of 4.15m/s.
The values I used for the period is .002sec so, a node will happen at half a period which is .001sec.
One wavelength is .325m and a node will happen half way through. This means that a node will occur at x=.1625m.
When I plug in the values in the above equation I get -.020599061m/s which is not the correct an answer. Would you please let me know if I am thinking about this incorrectly? Doesn't maximum velocity happen at the intersection point with the x-axis? Isn't this point a node in this case? This is an example in a textbook and I am trying to figure the maximum velocity by not just looking at the amplitude of the function.