ACPW help (1 Viewer)

Jkitty · Jul 4, 2015

quick question,
the velocity of a particle moving in SHM in a straight line is given by v^2=4x-x^2
c) find maximum speed of particle
how do you find maximum speed?

InteGrand · Jul 4, 2015

Jkitty said:
quick question,
the velocity of a particle moving in SHM in a straight line is given by v^2=4x-x^2
c) find maximum speed of particle
how do you find maximum speed?

$We are given $\text{speed}=|v|=\sqrt{4x-x^2}$. The quadratic under the square root in the R.H.S. is maximised at its vertex, which is at $x=\frac{-4}{2\times (-1)}=2$, so the maximal speed is $|v|_\text{max.}=\sqrt{4\times 2 - 2^2}=\sqrt{4}=2$.$

leehuan · Jul 5, 2015

Interesting way to solve it using x=-b/2a

At the same time, I'm looking at the question though and noticing that it's a part (c). Surely part (a) and part (b) would've helped somehow. This is a purely SHM method:

The endpoints of motion are found by solving for v=0. So we have 0^2 = 4x - x^2, solved to be x = 0, 4.
Hence the centre of motion is at x=2, which is the same as using x=-b/2a.
Because velocity is maximised at the centre of motion for SHM, sub that in to |v|

kawaiipotato · Jul 5, 2015

You could use inequalities, which is pretty fast
$v^2=4x-x^2$

$v^2 = - ((x-2)^2 - 4)$

$v^2 = 4 - (x-2)^2$

$(x-2)^2 \geq 0$

$-(x-2)^2 \leq 0$

$4 - (x-2)^2 \leq 4$

$v^2 \leq 4$

$-2 \leq v \leq 2$

$$ Hence, maximum v = 4$

Jkitty · Jul 7, 2015

There is another question for projectile motion,

25. a ball is thrown up with a velocity of 15 ms at an angle of 60°. At the same time, another ball is thrown towards the first one, at an angle of 45° and a velocity of 20 ms. The two balls are 30 m apart.
a. how far apart will they land?

kawaiipotato · Jul 7, 2015

$$ Ball A $: v = 15, \theta = 60$

$x'' = 0$ . $y'' = -10$

$x' = vcos\theta = 15cos60$ . $y' = -10t + vsin\theta = -10t + 15sin60$

$x = 15tcos60$ . $y = -5t^2 + 15tsin60$

$Let y = 0, for time at which ball A lands$

$\therefore t = 3sin60$

$$ Hence, $ x = 15(3sin60)(cos60)$

$x = \frac{45\sqrt3}{4}$

$$ Ball B $: v = 20, \theta = 180 - 45 = 135 ($ Shared origin, the angle is with respect to positive t-axis$

$x'' = 0$ . $y'' = -10$

$x' = vcos\theta = 20cos135$ . $y' = -10t + vsin\theta = -10t + 20sin135$

$x = 20tcos135 + 30$ . $y = -5t^2 + 20tsin135$

$Let $ y = 0 $ for time at which particle B lands$

$\therefore t = 4sin135$

$$ Hence, $ x = 20(4sin135)(cos135) + 30$

$x = -10$

$\therefore, $ They are $ \frac{45\sqrt3}{4} - (-10) = \frac{45\sqrt3}{4} +$ 10 metres apart. $$

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ACPW help (1 Viewer)

Jkitty

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InteGrand

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leehuan

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kawaiipotato

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Jkitty

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kawaiipotato

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