b) correct, the period would be 2π/60π = 1/30
c) when t = 3
d = (1/12)sin (180π) = 0
d) sin(60πt) = 0 when
t = 0 , 1/30, 1/15 ...
so the first positive value is 1/30
Given the equation for simple harmonic method d=1/12 sin(60 pi t) answer the following questions:
b)Find the frequency
i got 60pi
c)Calculate the value of d when t=3
d)Determine the least positive value of t for which d=0.
1 answer
2 answers