general equation
x = a sin(b(t-c))+d
Since equilibrium is x=0, d=0
x = a sin(b(t-c))
the period is 9, so b = 2π/9
x = a sin(2π/9 (t-c))
If it travels .4 m in a cycle, the amplitude is 0.1, so
x = 0.1 sin(2π/9 (t-c))
Now we have a problem. If x(0) = -0.1, that is its minimum value, so the velocity must be zero, not negative.
Check for a typo somewhere.
Recall that cos(0) is a max, so we can rewrite the above equation as
x = -0.1 cos(2π/9 t)
A particle moving in simple harmonic motion passes through the equilibrium point (x=0) 9 times per second. At t=0t=0 its velocity at x=−0.01 m is negative. It travels 0.4 m in a complete cycle.
The particle's position as a function of time is described by the following function:
x(t) = ___sin(___t +___) cm
please help
1 answer
1 answer