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March 27, 2017

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A 48-cm-long wire with a mass of 11.0 g is under a tension of 50.5 N. Both ends of the wire are held rigidly while it is plucked.

(a) What is the speed of the waves on the wire?
(b) What is the fundamental frequency of the standing wave?

The answers are 50.5 m/s and 48.9 Hz respectively. How do I get these answers?

  • Physics - ,

    For part (a) I've been doing the following work:

    u=M/L; where u= mass per unit length
    u=(0.011)/(.48)= 0.022916667

    v= sqrt(T/u); where T = tension
    v= sqrt(50.5/0.022916667) = 47.0 m/s

  • Physics - ,

    I'm not really sure how to figure out part (b); however, I'm sure that it's answer relies on me figure out part (a) correctly.

  • Physics - ,

    Your answer for part A is correct.
    Velocity in the stretched string is
    v = sqrt(T/m(o)).
    =sqrt(50.5•0.48/0.011)=46.94 m/s

    Part B

    Part B
    λ =2 L= 0.96 m.
    λ =v/f,
    f=v/ λ =46.94/0.96=48.9 m

  • Physics - ,

    Thank you. I guess WebAssign made a mistake.

  • Physics - ,

    Also, shouldn't f=48.9 m be f=48.9 Hz?

  • Physics - ,

    Certainly, Hz

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