Solve Wave Motion Question HRW 8 Ch 16 Q19

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SUMMARY

The discussion centers on solving a wave motion problem from HRW 8 Chapter 16 Question 19. The user calculated the angle X to be 19.47 degrees using the sine inverse function, resulting in a tension (T) of 20.79 N. The tension in the middle section (T2) was determined to be 6.93 N, and the mass per unit length was calculated as 6.67 x 10^-3 kg/m. The wave speed was computed to be 32.2 m/s, which slightly differs from the book's answer of 32.9 m/s, likely due to rounding errors.

PREREQUISITES
  • Understanding of wave motion principles
  • Knowledge of trigonometric functions, specifically sine and cosine
  • Familiarity with tension in strings and mass per unit length calculations
  • Ability to apply the wave speed formula v = √(T/μ)
NEXT STEPS
  • Review the derivation of wave speed in strings using Tension and mass per unit length
  • Explore the impact of rounding errors in physics calculations
  • Study the application of trigonometric functions in resolving forces
  • Investigate the differences in wave speed calculations in various mediums
USEFUL FOR

Students studying wave motion, physics educators, and anyone interested in the application of tension and wave speed in string dynamics.

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From HRW 8 Ch 16 Q19
http://img684.imageshack.us/img684/8947/82287578.jpg

What i did:
I resolved the components of tension is each string (T cosX and T sin X) and found X to be 19.47 degrees (sin inverse of [0.25m/0.75 m]). Then i got the value of T as 20.79 N. The Tension in the middle(between A and B) i called T2 and got it as 6.93 N [T sinX= T2]
Then, mass per unit length was found out to be 0.01kg/AB(1.5m) = 6.67x10^-3 kg/m
using the formula v= root of[T/mass per unit length], v= 32.2 m/s
The answer at the back of the book says 32.9 m/s .
Can anyone help me out here ?
Thanks
 
Last edited by a moderator:
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Probably just a rounding error on the book's part.
 

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