Refer to attachment. Please step it out for me
J jstout New member Joined Sep 2, 2009 Messages 1 Sep 2, 2009 #1 Refer to attachment. Please step it out for me Attachments question 9.jpg 73.8 KB · Views: 84
W wjm11 Senior Member Joined Nov 13, 2004 Messages 1,417 Sep 2, 2009 #2 The "simple" answer is that you take the derivative with respect to each of the four variables named. For an in depth review of vibrating string problems, see the 87 papers cited here: http://math.fullerton.edu/mathews/n2003 ... lnk_3.html That should pretty well cover it. ;-)
The "simple" answer is that you take the derivative with respect to each of the four variables named. For an in depth review of vibrating string problems, see the 87 papers cited here: http://math.fullerton.edu/mathews/n2003 ... lnk_3.html That should pretty well cover it. ;-)
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Sep 2, 2009 #3 I′ll do the rate of change of frequency when the force (T) is varied.\displaystyle I'll \ do \ the \ rate \ of \ change \ of \ frequency \ when \ the \ force \ (T) \ is \ varied.I′ll do the rate of change of frequency when the force (T) is varied. Let n = f(T) = 1DLgTπσ\displaystyle Let \ n \ = \ f(T) \ = \ \frac{1}{DL} \sqrt \frac{gT}{\pi\sigma}Let n = f(T) = DL1πσgT dndT = 1DLgπσ 12 (T)−1/2 = g2DL(πσT)\displaystyle \frac{dn}{dT} \ = \ \frac{1}{DL} \sqrt \frac{g}{\pi\sigma} \ \frac{1}{2} \ (T)^{-1/2} \ = \ \frac{\sqrt g}{2DL\sqrt( \pi \sigma T)}dTdn = DL1πσg 21 (T)−1/2 = 2DL(πσT)g
I′ll do the rate of change of frequency when the force (T) is varied.\displaystyle I'll \ do \ the \ rate \ of \ change \ of \ frequency \ when \ the \ force \ (T) \ is \ varied.I′ll do the rate of change of frequency when the force (T) is varied. Let n = f(T) = 1DLgTπσ\displaystyle Let \ n \ = \ f(T) \ = \ \frac{1}{DL} \sqrt \frac{gT}{\pi\sigma}Let n = f(T) = DL1πσgT dndT = 1DLgπσ 12 (T)−1/2 = g2DL(πσT)\displaystyle \frac{dn}{dT} \ = \ \frac{1}{DL} \sqrt \frac{g}{\pi\sigma} \ \frac{1}{2} \ (T)^{-1/2} \ = \ \frac{\sqrt g}{2DL\sqrt( \pi \sigma T)}dTdn = DL1πσg 21 (T)−1/2 = 2DL(πσT)g
