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InterviewSolution
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A long string having a cross- sectional area `0.80 mm^2` and density `12.5 g cm^(-3)` is subjected to a tension of 64 N along the X-axis. One end of this string is attached to a vibrator moving in transverse direction at a frequency fo 20Hz. At t = 0. the source is at a maximujm displacement y= 1.0 cm. (a) Write the equation for 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 ?A. `10sqrt(2)pi cm//s`B. `40sqrt(2) pi cm//s`C. `30sqrt(2)pi cm//s`D. `20sqrt(2) pi cm//s` |
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Answer» Mass per unit length of the string is `mu=Ad=(0.80 mm^(2))xx(12.5 g//cm^(3))=0.01 kg//m` `=(0.80xx10^(-6)m^(2))xx(12.5xx10^(3))=0.01 kg//m` speed of transverse waves produced in the string `v=sqrt((T)/(mu))=sqrt((64)/(0.01 kg//m))=80 m//s` the amplitude of the source is `a=1.0 cm` and the frequency is `f=20 Hz`. the angular frequency is `omega=2pif=40 pi//s`. also at `t=0`, the displacement is equal to its amplitute, `i.e.`, at `t=0, y=a`. the equation of motion of the source is, therefore, `y=(1.0 cm)cos[(40pis^(-1))t]` ...`(1)` the equation of the wave travelling on the string along the positive `X-`axis is obtained by replacing `t` by `[t-(x//v)]` in Eq. (i). It is, therefore, `y=(1.0 cm)cos[940pis^(-1)){t-(x//v)}x]` `=(1.0 cm)cos[(40pis^(-1))t-{(pi//2)m^-1)}x]` the displacement of the particle at `x=50 cm` at time `t=0.05 s` is obtained from Eq. (ii). `y=(1.0 cm)cos[(40pis^(-1))(0.05s)` `-{(pi//2)m^(-1}(0.5 m)]` `=(1.0 cm)cos[2pi-(pi//4)]` `=1.0 cm//sqrt(2)=0.71 cm` the velocity of the particle at position `x` at time `t` is also obtained from Eq. (ii). `V=(dely)/(delt)=-(1.0 cm)(40pi s^(-1))t-{(pi//2)m^(-1)}x]` `=-(40pi(cm)/(s))sin(2pi-(pi)/(4))` `=-(40pi)/sqrt(2) cm//s=-89 cm//s` |
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