Asked by Lewis
The displacement, s cm, of the end of a stiff spring at time, t sec, is given by s = ae^kt sin(2𝝅ft). Determine the velocity of the end of the spring after 2.5 secs if a= 6m, k= 0.55 and f = 15Hz
Answers
Answered by
mathhelper
s = ae^(kt) sin(2πft) ,subbing in the given:
s = 6e^(.55t) sin(2π(15)t)
v = ds/dt = [6e^(.55t)](cos(2π(15)t) )(30π) + [sin(2π(15)t)] (3.3e^(.55t) )
when t = 2.5
v = 6e^1.375 * 30π*cos(75π) + sin(75π)*3.3e^1.375
= ......
you do the button-pushing
check my typing, I used the product rule
s = 6e^(.55t) sin(2π(15)t)
v = ds/dt = [6e^(.55t)](cos(2π(15)t) )(30π) + [sin(2π(15)t)] (3.3e^(.55t) )
when t = 2.5
v = 6e^1.375 * 30π*cos(75π) + sin(75π)*3.3e^1.375
= ......
you do the button-pushing
check my typing, I used the product rule
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