Tension required to tune a string to the note of ##E_2##

AI Thread Summary
The discussion focuses on determining the tension required to tune a guitar string to the note E2, which has a frequency of 82.4 Hz. The boundary condition for a fixed string is that the displacement must be zero at both ends. The standing wave solutions are derived from the wave equation, leading to allowed oscillation frequencies expressed as f = (n/2L)√(W/M). The tension W can be calculated using the rearranged formula W = (4L²f²/n²)M, but the value of n needs to be clarified. It is suggested to investigate whether E2 corresponds to the fundamental frequency or a harmonic to determine the appropriate value for n.
thomas19981

Homework Statement


State the boundary condition which must be met at a point where the string of question 2 is fixed.

Hence find the real standing wave solutions to the wave equation, and determine the allowed oscillation frequencies, when such a string of length ##L## is fixed at its ends.

If the bottom string of a guitar has a mass of ##5.4 g/m## and its length is determined by the distance ##0.648 m## from the bridge to the nut, find the tension required to tune the string to the note known as ##E_2## (a frequency of ##82.4 Hz##)."

Question 2 describes a wave on a guitar string with the wave equation:

##\frac {\partial^2 \psi}{\partial t^2}=\frac WM \frac {\partial^2 \psi}{\partial x^2}##

Where $W$ is the tension, ##M## is the mass per unit length and ##\psi## is the displacement. It has been shown that the velocity of the wave is equal to ##\sqrt{W/M}##

Homework Equations

The Attempt at a Solution



So for the first part I stated that ##\psi## must be ##0## at the points where the sting is fixed as it has no displacement.

Next I stated that the real standing wave solution is given as follows: ##\psi=\psi_0cos(kx+\phi_x)cos(\omega t+\phi_t)##.

For the allowed oscillation frequencies I did the following:

##\psi=0## at ##x=0## and ##x=L##

Hence, ##cos(kx+\phi_t)=0## at these two points.

Setting ##x=0## yields that ##\phi_x=\pi/2##

Setting ##x=L## gives ##k=n\pi/L##

Since ##k## is the wavenumber ##\lambda=2\pi/k## so ##\lambda=2L/n##

##v=f\lambda## gives the allowed oscillation frequencies as ##f= \frac{n}{2L}\sqrt{W/M}##.

When I come to try to calculate the tension however I rearrange f above to get: ##W=\frac{4L^2f^2}{n^2}M## and I've been given all the numbers to calculate ##W## except I don't know what to pick for the value of ##n##. If I had to guess I would say ##n=2## just because in the question it asks for ##E_2## but I have no clue.

Any explanation or help would be very much appreciated.

 
Physics news on Phys.org
You need to determine whether the frequency known as E2 is the fundamental or lowest frequency for that guitar string. If it is, then ##n=1##. I suggest that you search "guitar tuning" on the web for clues to the answer.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top