Determine Tension and Length of 50g Guitar String Oscillates at 1280Hz

  • Thread starter Thread starter Ry122
  • Start date Start date
  • Tags Tags
    String Waves
AI Thread Summary
The discussion focuses on calculating the tension and length of a 50g guitar string oscillating at a frequency of 1280Hz, with a wavelength of 30cm. The calculated string length is determined to be 45cm, leading to a mass per unit length of 0.111 kg/m. Using the wave equation, the velocity is found to be 384 m/s, which leads to a tension calculation of approximately 16,367.616 N. Concerns are raised about the tension value being excessively high for a 50g string, prompting a search for potential errors in the calculations. The participants express confusion over the plausibility of the original question given the resulting tension.
Ry122
Messages
563
Reaction score
2
a third harmonic of a 50g guitar string oscillates at a frequency of 1280hz. the wavelength is 30cm
determine the tension in the string and the string length.

my attempt:
the 3rd harmonic of a guitar string is 3/2 times the wavelength so the length of the string is 45cm.
so mass per unit length is .05/.45=.111kg/m
v=f(lambda)
so v =384ms^1
V=sqrt(T/mu)
so 384=sqrt(T/.111)
T therefore is =16367.616N
isnt this value for tension too large to be correct?
where am i going wrong?
 
Physics news on Phys.org
I can't spot an error, the original question sounds plausible
and yet this tension is ridiculously high.
50g sounds a bit heavy, but even so...
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top