Interesting Case Of Oscillations

Relevant wiki: Hooke's Law

At time $t$ let the position of the mass be $x$ , then the extension of the left spring is $x$ . Because the right end of the right spring is constrained to be at $ut$ , the extension of the right spring is given by $ut-x$ . We therefore have $F=k(ut-2x)$

$a=\frac { { d }^{ 2 }x }{ { dt }^{ 2 } }$ .

Because the $ut$ term vanishes under two time derivatives, we can write $k(ut-2x)=\frac { -m }{ 2 } \frac { { d }^{ 2 }(ut-2x) }{ { dt }^{ 2 } }$ and thus $\frac { { d }^{ 2 }(ut-2x) }{ { dt }^{ 2 } } =-\frac { 2k }{ m } (ut-2x)$

Let $\gamma = ut-2x$ . In this form, we can see clearly that this is the governing equation of the simple harmonic oscillator, $\frac{d^2}{dt^2}\gamma = -\frac{2k}{m}\gamma$ , so that $\gamma = A\sin(\omega t+\phi)$ , or $2x= ut-A\sin(\omega t+\phi)$ At $t=o$ , $x=0$ , and therefore $\phi=0$ . Taking a time derivative, we have $u-2v=A\omega cos(\omega t)$ , this shows that $u-2v$ will be maximum for $v=0$ , which gives $A=u/\omega$

For $v=u$ , $cos(\omega t)=-1 \Rightarrow t=\pi /\omega =\pi \sqrt { \frac { m }{ 2k } }$

Similarly, we can find the distance which comes $D=\frac { \pi u }{ 2 } \sqrt { \frac { m }{ 2k } }$