Transverse waves are sent along a 5.00-m-long string with a speed of 30.00 m/s. The string is under a tension of 10.00 N. What i
s the mass of the string
1 answer:
Answer:
0.055 kg
Explanation:
Given that
Length of the string, l = 5 m
Speed of the wave, v = 30 m/s
Tension on the string, F(t) = 10N
From the formula written in the attachment, we have
v = velocity of the wave, in m/s
F(t) = Tension on the string, in N
U = Mass per length of the string, in kg/m
m = Mass of the string, in kg
l = Length of the string, in m
See attachment for the calculation
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Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s
By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s
Position #2:
Radius, r₂ = 2 ft
By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²
Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s
At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf
The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.
The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb
Answer: 44.25 ft-lb
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s
By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s
Position #2:
Radius, r₂ = 2 ft
By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²
Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s
At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf
The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.
The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb
Answer: 44.25 ft-lb