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    Physics for Scientists and Engineers Study Set 1
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    Exam 16: Superposition and Standing Waves
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    A String with Length L Is Fixed on Both Ends\(\lambda\)
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A String with Length L Is Fixed on Both Ends λ\lambdaλ

Question 16

Question 16

Multiple Choice

A string with length L is fixed on both ends. If λ\lambdaλ o = 2L and fo = v/ λ\lambdaλ 0, the wave function for the harmonic shown is  A string with length L is fixed on both ends. If  \lambda <sub>o</sub> = 2L and f<sub>o</sub> = v/ \lambda <sub>0</sub>, the wave function for the harmonic shown is   A)    B)    C)    D)    E)


A)  A string with length L is fixed on both ends. If  \lambda <sub>o</sub> = 2L and f<sub>o</sub> = v/ \lambda <sub>0</sub>, the wave function for the harmonic shown is   A)    B)    C)    D)    E)
B)  A string with length L is fixed on both ends. If  \lambda <sub>o</sub> = 2L and f<sub>o</sub> = v/ \lambda <sub>0</sub>, the wave function for the harmonic shown is   A)    B)    C)    D)    E)
C)  A string with length L is fixed on both ends. If  \lambda <sub>o</sub> = 2L and f<sub>o</sub> = v/ \lambda <sub>0</sub>, the wave function for the harmonic shown is   A)    B)    C)    D)    E)
D)  A string with length L is fixed on both ends. If  \lambda <sub>o</sub> = 2L and f<sub>o</sub> = v/ \lambda <sub>0</sub>, the wave function for the harmonic shown is   A)    B)    C)    D)    E)
E)  A string with length L is fixed on both ends. If  \lambda <sub>o</sub> = 2L and f<sub>o</sub> = v/ \lambda <sub>0</sub>, the wave function for the harmonic shown is   A)    B)    C)    D)    E)

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