Question

Consider a harmonic wave with the following wave function: y=(12cm)cos(?4.5cmx??16st).

What is the least amount of time required for a given point on this wave to move from

y = 0 to y = 12cm?
Express your answer using two significant figures.

t = s   

Answer 1

Compare the given equation with standard equation:

y = A*cos (kx - wt)

w = 16*pi

Time period of wave will be

T = 2*pi/w = 2*pi/(16*pi) = 0.125 sec

Now Amplitude of wave = 12 cm

we need minimum time required from y = 0 to y = 12 cm

y = 0 is equilibrium point and y = 12 cm is max amplitude

So time required from equilibrium point to max amplitude = T/4 = 0.125/4 = 0.031 sec

Please Upvote.

(Time T = equilibrium to max amplitude + than back to equilibrium + than to opposite side max amplitude + than return back to equilibrium point).

in your given equation if w = pi/16, then T = 2*pi/w = 2*pi/(pi/16) = 32 sec and required time = T/4 = 32/4 = 8 sec