Answer :
Answer:
a) 342 m/s
b) 51*10^-6 m/s
c) 0.87m/s^2
Explanation:
The following function describes the displacement of the molecules in a sound wave:
[tex]s(x,t)=3.00nm\ cos(50.00\ m^{-1}x-1.71*10^4s^{-1}t)[/tex] (1)
The general form of a function that describes the same situation is:
[tex]s(x,t)=Acos(kx-\omega t)[/tex] (2)
By comparing equations (1) and (2) you have:
k: wave number = 50.00 m^-1
w: angular frequency = 1.71*10^4 s^-1
A: amplitude of the oscillation = 3.00nm
a) The speed of the sound is obtained by using the formula:
[tex]v=\frac{\omega}{k}=\frac{1.71*10^4s^-1}{50.00m^{-1}}=342\frac{m}{s}[/tex]
b) The maximum speed of the molecules is the maximum value of the derivative of s(x,t), in time. Then, you first obtain the derivative:
[tex]\frac{ds}{st}=-\omega A sin(kx-\omega t)[/tex]
The max value is:
[tex]v_{max}=\omega A[/tex]
[tex]v_{max}=(1.71*10^4s^-1)(3.00nm)=51300\frac{nm}{s}=51\frac{\mu m}{s}[/tex] = 51*10^-6 m/s
c) The acceleration is the max value of the derivative of the speed, that is, the second derivative of the displacement s(x,t):
[tex]a=\frac{dv}{dt}=\frac{d^2s}{dt^2}=-\omega^2A cos(kx-\omega t)\\\\a_{max}=\omega^2 A[/tex]
Then, the maximum acceleration is:
[tex]a_{max}=(1.71*10^4s^{-1})^2(3.00nm)=0.87\frac{m}{s^2}[/tex]