Physics Traveling Wave?
The position of a particle is given by x = (8.0 cm) sin 4πt, where t is in seconds.
Determine the frequency of the particle's motion.
Determine the period of the particle's motion.
Determine the amplitude of the particle's motion.
What is the first time after t = 0 that the particle is at its equilibrium position?
Answers & Comments
Verified answer
The given equation can be compared to the formula x = A sin Wt; where A is the amplitude and W is the angular velocity.
A1) W = 2 TT n (where n is the frequency)
So comparing your equation with the above equation we can say that 4 TT is the value for W.
Therefore, 4 TT = 2 TT n
4 TT / 2 TT = n
n = 2.
A2) Time period is the inverse of frequency so n = 1/T
1/2 = T
Therefore T = 0.5 sec
A3) Amplitude should be 8. (we get that by comparing it with the above equation)
A4) Since the time period is 0.5 sec, the particle should return to its equilibrium position for the first time at T/2. T/2 = 0.25 sec
1) w= 2 TT n
here , 2 TT n = 4 TT
therefore, n= 2
2) T = 1/n = 1/2 = 0.5 sec
3) a =8
as y = x sin w t
where y -- displacement from equillibrium position ,,, x- amplitude
4) it will return to equillibrium at T /2 = 0.25 sec after t=0..... because at this point it will be lambda/2 and particl;e will try to cover second half of wave.