Question

Ignore air resistance. Consider a knuckleball with lateral motion satisfying the initial value problem x’’(t) = -25 sin (4ωt + θ0), x’(0) = x(0) = 0. Whit θ0 = 0 and ω = 1, find an equation for x(t) and graph the solution for 0 ≤ t ≤ 0.68. (HINT: Use integration by subtitution)

Answer 1

\[x''(t)=-25\sin(4\omega t+\theta_0)\]
With the initial conditions \(x'(0)=x(0)=0\). Also, you're given actual values of \(\omega\) and \(\theta_0\), so you have the differential equation
\[x''(t)=-25\sin(4t)\]
Integrate both side with respect to \(t\):
\[\int x''(t)~dt=\int-25\sin(4t)~dt\\
x'(t)=-25\left(\cdots\right)+C_1\]
Integrate again:
\[\int x'(t)~dt=\int\left[-25\left(\cdots\right)+C_1\right]~dt\\
x(t)=\cdots\]
I'll let you fill in the blanks.

Your general solution will have some constants, \(C_1\) and \(C_2\). You solve for these using the initial values.

Answer 2

ok

Answer 3

would that be 16 in the blanks?

Answer 4

@SithsAndGiggles

Answer 5

@SithsAndGiggles

Answer 6

\[\begin{align*}x'(t)&=\int(-25\sin(4t))~dt\\
&=-25\int\sin(4t)~dt\\
&=-25\left(-\frac{1}{4}\cos(4t)+C_1\right)\\
&=\frac{25}{4}\cos(4t)+C_1
\end{align*}\]
Integrate both sides again, and you have \(x(t)=\cdots\).

Now, use \(x'(0)=0\):
\[x'(0)=0=\frac{25}{4}\cos(4\cdot0)+C_1~\Rightarrow~C_1=-\frac{25}{4}\]
Do the same with \(x(0)=0\) to solve for \(C_2\) in \(x(t)=\cdots\).

Answer 7

@rizwan_uet can u help?

Answer 8

i already did what sith told me and i got x(t) = 25/16 sin(4t) for the first part..but i dont get when he said do the same with x(0) = 0 to solve for c2 in (t)

Answer 9

ok let me check

Answer 10

so what you got when you integrated it again???

Answer 11

i got 25/16 sin(4t) + c1

Answer 12

isnt it become
x(t)= 25/16 sin(4t) + c2 ????? you put c1 here but its c2 actualy

Answer 13

ohhh ok

Answer 14

puting x(0) = 0 in this equation now you can get c2
thus 0 = 25/16sin(4x0) +c2

Answer 15

ohhh okay i see.

Answer 16

thanks

Answer 17

you are welcome