Help wit Q - (1 Viewer)

hihihi · Jul 15, 2016

A ball on the end of a spring moves according to ¨a = −256x (in units of centimetres
and minutes). The ball is pulled down 2cm from the origin and released. Find the
speed at the centre of motion.

Thx

InteGrand · Jul 15, 2016

hihihi said:
A ball on the end of a spring moves according to ¨a = −256x (in units of centimetres
and minutes). The ball is pulled down 2cm from the origin and released. Find the
speed at the centre of motion.

Thx

$\noindent Note the angular frequency parameter satisfies $n^2 = 256\Rightarrow n = 16$. Now, clearly the amplitude is $A = 2$ (because we pull the mass down by 2 cm and then just release), so using $v^2 = n^2\left(A^2 -x^2\right)$, the speed at the centre of motion satisfies $v^2 = n^2 A^2$ (setting $x=0$), whence $v = nA = 16\times 2 = 32$. I.e. answer is 32 centimetres per minute.$

HeroicPandas · Jul 15, 2016

(Another method)

The ball being pulled down 2cm from the origin implies that v = 0 at x = 2 or v(2) = 0. So we need an equation relating v and x.

Remember that $a = -n^2 x$ , which means $n= 16$ .

By chain rule,

$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v$

Now, we have

$\frac{dv}{dx} \cdot v = -256x$

$v dv = -256x dx$

Integrating both sides to arrive at

$\frac{v^2}{2} = -128x^2 +C$

Use the initial condition (v = 0 at x = 2) to find that $C = 512.$

So,

$v^2 = -256x^2 + 1024$

At extreme points, $v = 0$ and solving for x gives $x = \pm 2$ . So centre of motion is at $x = 0$ hence velocity at centre of motion is $v = \sqrt{1024} = \pm 32$ .

ProdigyInspired · Jul 15, 2016

HeroicPandas said:
(Another method)
$a = -n^2 x$ , which means $n= 16$ .

I thought this is 3U content (Simple Harmonic Motion)?
Is $a = -n^2 x$ taught in 2U?

Unless this is a 3U question?

InteGrand · Jul 15, 2016

ProdigyInspired said:
I thought this is 3U content (Simple Harmonic Motion)?
Is $a = -n^2 x$ taught in 2U?

Unless this is a 3U question?

Yeah it's 3U content (not 2U). It should be a 3U question.

Help wit Q - (1 Viewer)

hihihi

Member

InteGrand

Well-Known Member

HeroicPandas

Heroic!

ProdigyInspired

Tafe Advocate

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)