A standing wave pattern on a string is described by y(x, t) = 0.040 sin (5 pi x)

Question

A standing wave pattern on a string is described by y(x, t) = 0.040 sin (5 pi x)(cos 40 pi t), where x and y are in meters and t is in seconds. For x greaterthanorequalto 0, what is the location of the node with the (a) smallest, (b) second smallest, and (c) third smallest value of x? (d) What is the period of the oscillatory motion of any (nonnode) point? What are the (e) speed and (f) amplitude of the two traveling waves that interfere to produce this wave? For t greaterthanorequalto 0, what are the (g) first, (h) second, and (i) third time that all points on the string have zero transverse velocity? (a) Number Units (b) Number Units (c) Number Units (d) Number Units (e) Number Units (f) Number Units (g) Number Units (h) Number Units (i) Number Units

Explanation / Answer

A) Node points are at lamda/2 ,lamda,3*lamda/2....

so the answer for a) at x = lamda/2

comparing the given equation with general equation

y(x,t) = 0.040*sin(5*pi*x)cos(40*pi*t)

general equation is

y(x,t) = 2A*sin(kx)cos(wt)

k = 2*pi/lamda

lamda = 2*pi/k = (2*3.142)/(5*3.142) = 0.4 m

so answer for A) is lamda/2 = 0.4/2 = 0.2 m

b) second smallest is at lamda = 0.4 m

c) third smallest is at 3*lamda/2 = 3*0.2 = 0.6 m

d) T = 2*pi/w = (2*3.142)/(40*3.142) = 0.05 sec

e) speed is v = w/k = 40pi/5pi = 8 m/sec

f) amplitude is A = 0.04/2 = 0.02 m

g) zero transverse velocity is at anti node point

So first node point will be at T/4 = 0.05/4 = 0.0125 sec

h) t = (T/2)+(T/4) = 3*T/4 = 3*0.05/4 = 0.0375 sec

i) t= T+(T/4) = 0.05+(0.0125) = 0.0625 sec

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