A mass stretches a spring by 10 cm. The mass is released from the equilibrium position with an upwards starting velocity of $v$ cm/s, where $v > 0$ is a given constant. There is no damping. In this problem, for simplicity use $g = 1000$ cm/s^2
a.) Find the frequency, period, amplitude, and phase of the resulting periodic motion, in terms of $v$.
b.) Suppose that we don't want the mass to ever be more than 5 cm below the equilibrium position. What values of $v$ may we use?
c.) Suppose we want the mass to be 5 cm below the equilibrium position after 10 seconds. What values of $v$ may we use? What is the smallest value of $v$ that works?
I wish to mostly focus on part c.), however, I will provide my work for part a.) and b.), just in case I made a mistake.
a.) $\omega = \sqrt\frac{k}{m}=\sqrt\frac{g}{L}=10$ 1/s $$u(t) = A\cos10t +B\sin10t$$ $u(0) = 0 \rightarrow A = 0 $ $$u'(t) = -10A\sin10t +10B\cos10t$$ $u'(0) = 10B = v \rightarrow B = \frac{v}{10}$ $$u(t) = \frac{v}{10}\sin10t$$
Period = $\frac{2\pi}{\omega}=\frac{\pi}{5}$, Amplitude = $\sqrt{A^2+B^2}=B=\frac{v}{10}$, $\cos\delta = 0, \sin\delta = 1 \rightarrow \delta = \frac{\pi}{2}$
So therefore we have $$u(t) = \frac{v}{10}\cos{(10t-\frac{\pi}{2})}$$
b.) We want $u(t) \leq 5$, so we have $$\frac{v}{10}\cos{(10t-\frac{\pi}{2})} \leq \frac{v}{10} \leq 5 \rightarrow v \leq 50$$ Therefore any $0 < v \leq 50$ will suffice.
c.) Here is where I am confused, as the question asks for a range of values for $v$, but it is given at a precise time $t = 10$, so plugging in this value should give you an exact value for $v$ at the given time $t = 10$ for $u(t) = 5$ right? Am I misunderstanding some aspect of the problem here?
Note that $\sin(100) \approx -0.506 < 0$, so we need this inequality to hold
$$ u(10) = \frac{v}{10}\sin(100) \le -5 $$
$$ \implies \frac{v}{10}|\sin(100)| \ge 5 $$
$$ \implies v \ge \frac{50}{|\sin(100)|} \approx 98.7 $$