# Deriving a DE for an undamped mass spring system and analyzing its properties

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
$$u(10) = \frac{v}{10}\sin(100) \le -5$$
$$\implies \frac{v}{10}|\sin(100)| \ge 5$$
$$\implies v \ge \frac{50}{|\sin(100)|} \approx 98.7$$