A wire with mass 44.0g is stretched so that its ends are tied down at points a d

Question

A wire with mass 44.0g is stretched so that its ends are tied down at points a distance 77.0cm apart. The wire vibrates in its fundamental mode with frequency 55.0Hz and with an amplitude at the antinodes of 0.330cm .

Part A

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

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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.

Subm

A wire with mass 44.0g is stretched so that its ends are tied down at points a distance 77.0cm apart. The wire vibrates in its fundamental mode with frequency 55.0Hz and with an amplitude at the antinodes of 0.330cm .

Part A

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

v =   m/s  

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Part B

Compute the tension in the wire.

F =   N  

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Part C

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

vmax =   m/s  

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Part D

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

amax =   m/s2  

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Explanation / Answer

a) velocity = frequency*lambda

lambda = 2L = 2*0.77 = 1.54 m

v = 55*1.54 = 84.7 m/s

b) Velocity of wave = square root of (Tension/mass per unit length)

Tension = (Velocity of wave)^2 x mass per unit length

T = 84.7^2*(44x10^-3/0.77)

T = 409.948 N

c) maximum transverse velocity = A*w = 2pi*f*A

maximum transverse velocity = 2*pi*55*0.33x10^-2

maximum transverse velocity = 1.14 m/s

d) maximum acceleration = Aw^2 = A*(2pi*f)^2

maximum acceleration = 0.33x10^-2*(2pi*55)^2

maximum acceleration = 394.093 m/s^2

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