A spring hanging from the ceiling vibrates up and down. it's vertical position at time (t) is given by f(t)=4 sin(3t). (a) Find the velocity of the spring at time t. (b) What is the spring's maximum speed. (c) what is location when it reaches its maximum speed.
for a .. i took the first derivative which is 4 cos(3t).3 = 12 cos(3t)
and for (b) .. s it 12 ? =_=
;))
Help Me or im gonna Die thinking o(>< )o
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Me is too Dump to Solve this (⇀‸↼‶)7
vertical position is given which is distance.so f'(t)=4cos(3t)
it is velocity
o_o aha .. what about (b) ? =..= its killing me
Vmax = A*ω and ω = √(k/m).Maximum speed is found by this formula.
r u sure ? :l
k i will try but what us a and w ?
u r right.but we must have to solve it by some derivative method.i have just finished the preparation for my exam which is going to b held tomorrow.so now i m mentally tired.could not get any solution right at this time.sorry.i will tell u later if u could not solve it.but if u solve it then msg me plz.thanks n good night
:l
Yes its 12 for (b)
(c) it comes to its mean position when it reaches its maximum speed
(c) you get max speed when t=0 Substitute t=0 in f(t) and f(t) will become zero which shows that velocity is max at mean position
You could also find a(t) and find where that = 0 to find the max I think
okie .. :l
thanx ppl =_='
i think im gonna close this question :B .. can't solve it anyway
u still not satisfied by answers?
first derivative is 12cos(3t).which is velocity. and the maximum speed is 12 coz maximum value of cos is 1and if u put 1 in equation f't=12cos(3t) in place of cos(3t) u get maximum speed which is 12.now when u have maximum speed 12 and u know t=0 because cos(3t)=1 only when 3t=0 as cos0=1.so now put t=0 in first equation f(t)=4sin(3t) and you get location which is zero.that's ans of c.
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