A wire with mass 38.0 g is stretched so that its ends are tied down at points a

Question

A wire with mass 38.0 g is stretched so that its ends are tied down at points a distance 82.0 cm apart. The wire vibrates in its fundamental mode with frequency 55.0 Hz and with an amplitude at the antinodes of 0.340 cm .

Part A

What is the speed of propagation of transverse waves in the wire?

90.2

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Correct

Part B

Compute the tension in the wire.

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Part C

Find the magnitude of the maximum transverse velocity of particles in the wire.

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Part D

Find the magnitude of the maximum acceleration of particles in the wire.

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90.2

Explanation / Answer

m = 38 g, L = 82 cm

f = 55 Hz, A = 0.34 cm

(a) f = v/2L

55 = v/(2*0.82)

v = 90.2 m/s

(b) v = (F/u)^0.5

u is linear mass density u = m/L

90.2 = (F*0.82/0.038)^0.5

F = 377.04 N

(c) w = 2pi*f

vmax = Aw

vmax = 0.0034*2*3.14*55

vmax = 1.174 m/s

(d) amax = Aw^2

amax = 0.0034*(2*3.14*55)^2

amax = 405.6 m/s^2

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