Question #85455

. The equation of a stationary wave on a string fixed at both ends is given by
y( x,t) = 2sin πx cos100πt
where x and y are measured in metre and t in second. Calculate the amplitude,
wavelength and frequency of component waves whose superposition generated this
stationary wave. Also write the equations of component waves.

Expert's answer

Question # 85455, Physics / Mechanics | Relativity

Task: The equation of a stationary wave on a string fixed at both ends is given by y(x,t)=2sin(πx)cos(100πt)y(x,t)=2\sin(\pi x)\cos(100\pi t), where xx and yy are measured in metre and tt in second. Calculate the amplitude, wavelength and frequency of component waves whose superposition generated this stationary wave. Also write the equations of component waves.

Solution:

y(x,t)=2sin(πx)cos(100πt)=sin(πx+100πt)+sin(πx100πt)=sin(πx+100πt)+sin(100πtπx+π)y(x,t)=2\sin(\pi x)\cos(100\pi t)=\sin(\pi x+100\pi t)+\sin(\pi x-100\pi t)=\sin(\pi x+100\pi t)+\sin(100\pi t-\pi x+\pi)

General formula for a standing wave: y(x,t)=Asin(ωtkx+ϕ)y(x,t)=A\sin(\omega t-kx+\phi).

We have y(x,t)=y1(x,t)+y2(x,t)y(x,t)=y_{1}(x,t)+y_{2}(x,t).

Consider first component y1(x,t)=sin(πx+100πt)y_{1}(x,t)=\sin(\pi x+100\pi t):

amplitude is A1=1A_{1}=1 m

angular frequency is ω1=100π\omega_{1}=100\pi rad/s

frequency is f1=ω1/2π=50f_{1}=\omega_{1}/2\pi=50 Hz

wave number is k1=πk_{1}=-\pi rad/m

velocity is v1=ω1/k1=100v_{1}=\omega_{1}/k_{1}=-100 m/s

(velocity is negative because the wave is travelling in the negative xx direction)

wavelength is λ1=v1/f1=2\lambda_{1}=|v_{1}|/f_{1}=2 m

Consider second component y2(x,t)=sin(100πtπx+π)y_{2}(x,t)=\sin(100\pi t-\pi x+\pi):

amplitude is A2=1A_{2}=1 m

angular frequency is ω2=100π\omega_{2}=100\pi rad/s

frequency is f2=ω2/2π=50f_{2}=\omega_{2}/2\pi=50 Hz

wave number is k2=πk_{2}=\pi rad/m

velocity is v2=ω2/k2=100v_{2}=\omega_{2}/k_{2}=100 m/s

wavelength is λ2=v2/f2=2\lambda_{2}=|v_{2}|/f_{2}=2 m


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