The aluminum rod AB (G = 26 GPa) is bonded to the brass rod BD (G = 39 GPa). Knowing that portion CD of the brass rod is hollow
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A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched
<u>Explanation:</u>
Work, W = 2 J
Initial distance, = 30 cm
Final distance, = 42 cm
Force, F = 30 N
Stretched length, x = ?
We know,
W = 1/2 kΔx²
Δx = 42-30 cm = 12 cm = 0.12 m
2 J = 1/2 k X (0.12)²
k = 277.77 N/m
According to Hooke's law,
F = kx
30 N = 277.77 X x
x = 0.108 m
x = 10.8 cm
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.
(a) The lowest frequency (called fundamental frequency) of a wire stretched under a tension T is given by
where
L is the wire length
T is the tension
m is the wire mass
In our problem, L=10.9 m, m=55.8 g=0.0558 kg and T=253 N, therefore the fundamental frequency of the wire is
b) The frequency of the nth-harmonic for a standing wave in a wire is given by
where n is the order of the harmonic and f1 is the fundamental frequency. If we use n=2, we find the second lowest frequency of the wire:
c) Similarly, the third lowest frequency (third harmonic) is given by
where
L is the wire length
T is the tension
m is the wire mass
In our problem, L=10.9 m, m=55.8 g=0.0558 kg and T=253 N, therefore the fundamental frequency of the wire is
b) The frequency of the nth-harmonic for a standing wave in a wire is given by
where n is the order of the harmonic and f1 is the fundamental frequency. If we use n=2, we find the second lowest frequency of the wire:
c) Similarly, the third lowest frequency (third harmonic) is given by