Concept explainers
(a)
To Find:The maximum kinetic energy of the wire.
(a)
Explanation of Solution
Given:
Length of the wire, l=2.00 m
Tension in the wire, T=40.0 N
Mass of the wire, m=0.100 kg
At the midpoint, amplitude is A=2.00 cm = 0.02 m
Formula Used:
Maximum kinetic energy of the wire can be obtained by:
K.Emax=14mω2A2
Here, m is the mass, ω is the angular frequency and A is the amplitude of the wave.
ω=2πf
Here, f is the frequency which can be obtained by:
f=12l√Tμ
μ=ml
Calculations:
Find the mass per unit length:
μ=ml=0.100 kg2.00 m=0.05 kg/m
Now calculate the frequency of the vibrating wire in fundamental mode:
f=12l√Tμ=12×2.00√40.00.1/2.00=7 Hz
The angular frequency is:
ω=2πf=2π(7.00)≈44 rad/s
Now substitute all the known values to find the maximum kinetic energy of the wire:
K.Emax=14mω2A2=14(0.100)(44 rad/s)2(0.02m)2=0.0194 J=19.4 mJ
Conclusion:
Thus, the maximum kinetic energy of the wire is 19.4 mJ .
(b)
To Find: The kinetic energy of the wire at the instant when transverse displacement is given by y=0.0200sin(π2x) .
(b)
Explanation of Solution
Given:
Length of the wire, l=2.00 m
Tension in the wire, T=40.0 N
Mass of the wire, m=0.100 kg
At the midpoint, amplitude is A=2.00 cm = 0.02 m
Displacement, y=0.0200sin(π2x)
0.00 m≤x≤2.00 m
Formula Used:
Wave equation of standing wave in fundamental mode:
y=Asin(kx)cos(ωt)
Calculations:
Compare the given displacement and the wave equation:
y=0.0200sin(π2x) and y=Asin(kx)cos(ωt)
⇒cos(ωt)=1⇒ω=0∴K.E=0
Conclusion:
Thus, the kinetic energy at the given instant would be zero.
(c)
To Find: The value of x for which the average value of the kinetic energy per unit length is the greatest.
(c)
Explanation of Solution
Given:
Length of the wire, l=2.00 m
Tension in the wire, T=40.0 N
Mass of the wire, m=0.100 kg
At the midpoint, amplitude is A=2.00 cm = 0.02 m
Displacement, y=0.0200sin(π2x)
0.00 m≤x≤2.00 m
Formula Used:
Average value of kinetic energy per unit length:
dKdx=12μ(∂y∂t)2
Here, μ is the mass per unit length.
Wave equation of standing wave in fundamental mode:
y=Asin(kx)cos(ωt)
Calculations:
∂y∂t=Asinkx(−ωsinωt)
For maxima, equate the derivative with zero.
sin(kx)=0kx=0⇒x=0orkx=πx=π2π/λx=λ2=(2l)2x=2(2.00)2=2.00 m
Conclusion:
Thus, the value of x for which the average value of the kinetic energy per unit length is the greatest is 0.0 m & 2.00 m .
(d)
To Find: The value of x for which the elastic potential energy per unit length has the maximum value.
(d)
Explanation of Solution
Given:
Length of the wire, l=2.00 m
Tension in the wire, T=40.0 N
Mass of the wire, m=0.100 kg
At the midpoint, amplitude is A=2.00 cm = 0.02 m
Displacement, y=0.0200sin(π2x)
0.00 m≤x≤2.00 m
Formula Used:
Average value of elastic potential energy per unit length:
dUdx=12μ(∂y∂x)2
Here, μ is the mass per unit length.
Wave equation of standing wave in fundamental mode:
y=Asin(kx)cos(ωt)
Calculations:
∂y∂x=Akcoskxcosωt
For maxima, equate the derivative with zero.
cos(kx)=0kx=π2x=π/22π/λx=λ4=2l4x=2(2.00)4=1.0 m
Conclusion:
Thus, the value of x for which the average value of the elastic potential energy per unit length is the greatest is at 1.0 m .
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