1. A wire of length L is fixed at both ends. Which of the options in figure 1 re
ID: 2294535 • Letter: 1
Question
1. A wire of length L is fixed at both ends. Which of the options in figure 1 represents the stationary wave with the longest wavelength that can be established in this wire?
Answer:______
3. Which of the options in figure 1 corresponds to the second harmonic, i.e, the stationary wave with the second longest wavelength that can be established in this wire?
Answer:______
Answer:______
Let us assume that the speed of the wave in our string is v. Write an experience for f1, the frequency of the first harmonic as a function of L and v.
f1= <equation 1>
f1=
Equation 3 will be useful during the lab and to answer the next three questions
5. Two identical wires A and B are subject to tension. The tension in wire A is ___times larger than that in wire B. Find the ratio of the frequencies of the first harmonic in these two wires, fA1/fB1
6. Two wires A and B are subject to the same tension, but A has___ times the mass per cm than wire B, Find the ratio of the frequencies of the first harmonic in these two wires. fA1/fB1
7. Two wires A and B are subject to the same tesion and have the same mass per unit length. The length of wire A is______ times that of wireB. Find the ratio of the frequencies of the first harmonic in these two wires, fA1/fB1
Please provide all the answers, explanations would be better. Thank you so much!!!!!!!!!!!
Explanation / Answer
1) C
2) lamda1 = 2*L
3) A
4) lamda2 = L
f1 = (v/2*L) = sqrt(T/mue)/2*L
f2 = 2*V/(2*L) = sqrt(T/mue)/L
5)
if TA = n*TB
then, fA1/fB1 = sqrt(n)
6)
if mA = n*mB
then, fA1/fB1 = sqrt(1/n)
7)
if LA = n*LB
then, fA1/fB1 = n
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.