Question

y (x , t)= A sin (k x − ω t )
calculate the vertical position y in [cm], vertical speed vy in [m/s] and vertical acceleration ay in [m/s2] of the wave that the wave driver generated at x = 15 cm and at time t = 3 s during trial 5 of this experiment. Assume that A = 2 cm. The other values in the equation can be calculated from the experimental data. Remember to differentiate the above equation to get the expressions for vy and ay. Answers: y = 1.5 cm, v = - 3.9 m/s, a = -1500 m/s2

Trail 5 m=450g, f=49.0Hz(f^2=2401Hz)

Total string length	225cm
String length between pulley and wave driver	160.5cm
Total string mass	3.7g

Answer 1

given :

A = 2 cm = 0.02 m

m=450g = 0.45 kg

f=49.0Hz

x = 15 cm = 0.15m

t = 3 s

l = 225cm = 2.25 m

string mass = 3.7g = 0.0037 kg

wave equation : y (x , t)= A sin (k x − ω t )

Now, Tension = T = mg = 0.45 * 9.81 = 4.4145 N

mass per unit length of string = = string mass / length = 0.0037 / 2.25 = 0.0016 kg/m

wave speed = .

angular frequency =

angular wave number = .

Now, the wave equation can be written as y (x,t) = 0.02 m sin(5.86 x - 307.88 t)

at, t = 3 s and x = 0.15 m, displacement y =  0.02 m sin(5.86 *0.15 - 307.88*3) = +0.00774 m [answer]

on diferentia y(x,t) with respect to time, we velocity of particle (x,t)  = -(0.02 * 307.88) m/s cos ( 5.86 x - 307.88 t)

at t= 3s, x = 0.15 m, Velocity of particle = -(0.02 * 307.88) m/s cos (5.86 *0.15 - 307.88*3) = +5.678 m/s [answer]

on further differentiation with respect to t, we get acceleration = (0.02*307.88*307.88) m/s2 sin(5.86 x - 307.88 t).

at t= 3s, x = 0.15 m, acceelration of particle = = (0.02*307.88*307.88) sin(5.86 *0.15 - 307.88*3) = +733.463 m/s/s. [answer]