A string made of a material of density 12.5 g/cc has a cross-sectional area 0.80 mm?. This string is subjected to a tension of 64 N along the horizontal axis. One end of the string is attached to a vibrator moving in a transverse direction at a frequency of 20 Hz. At t=0, the source is at a maximum displacement of 1.0 cm. (a) Find the speed of the wave travelling on the string? (b) Write the equation of the wave. (c) What is the displacement of the particle of the string at x 50 cm at time t 0.05 s. (d) What is the velocity of this particle at this instant?
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Answer:
Mass per unit length of the string=μ=ρA=0.8×10
−6
×12.5×10
3
=0.01kg/m
Thus speed of the wave=
μ
T
=
0.01
64
=80m/s
Amplitude of the wave=A=1cm
ω=2πν=40πs
−1
v=
k
ω
⟹k=
80
40π
=
2
π
m
−1
Thus the wave equation is y=Acos(ωt−kx)
=(1cm)cos[(40πs
−1
)t−(
2
π
m
−1
)x]
Explanation:
Answered by
0
Answer:
ANSWER
Mass per unit length of the string=μ=ρA=0.8×10
−6
×12.5×10
3
=0.01kg/m
Thus speed of the wave=
μ
T
=
0.01
64
=80m/s
Amplitude of the wave=A=1cm
ω=2πν=40πs
−1
v=
k
ω
⟹k=
80
40π
=
2
π
m
−1
Thus the wave equation is y=Acos(ωt−kx)
=(1cm)cos[(40πs
−1
)t−(
2
π
m
−1
)x]
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