The displacement of the air molecules in a sound wave is modeled with the wave function s(x, t) = 3.00 nm cos(50.00 m−1x − 1.71 ✕ 104 s−1t). (a) What is the wave speed (in m/s) of the sound wave? 342 Correct: Your answer is correct. m/s (b) What is the maximum speed (in m/s) of the air molecules as they oscillate in simple harmonic motion? m/s (c) What is the magnitude of the maximum acceleration (in m/s2) of the air molecules as they oscillate in simple harmonic motion?

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:

 (1)

The general form of a function that describes the same situation is:

  (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:

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:

The max value is:

= 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):

Then, the maximum acceleration is:

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm