more questions on simple harmonic motion

**differentiate** · September 26th 2009, 04:36 AM

Could someone check these for me? i was told to put g = 9.8 m/s for all calculus questions, and I think it's a bit ridiculous.

1. A long jumper running at 8 m/s jumps into the air, rising to a height of 0.7 metres. What is the length of the jump.

maximum height
$0.7 = \frac{V^2sin^2\alpha }{2g}$

$0.7 = \frac{8^2sin^2\alpha }{2\cdot 9.8}$

$13.72 = 64sin^2\alpha$

$sin\alpha = 0.463006479$

$\alpha = 27^{\circ}{35}'$

Range
$\frac{V^2sin2\alpha }{9.8}$

$=\frac{ 64sin55^{\circ}{10}' }{9.8}$

$= 5.36 m$

This is wrong. the answer says 5.29 m.

2. The velocity of a point moving along the x axis is given by
$V^2 = 10 + 8x - 2x^2$
Show that it is simple harmonic.

$\ddot{x} = \frac{d}{dx}(\frac{1}{2}v^2)$

$= \frac{d}{dx}(\frac{1}{2}10 + 8x - 2x^2)$

$= \frac{d}{dx}5+4x - x^2$

$= 4 - 2x$

However this is not the form $-n^2x$

so how would I do this?

Thanks!!!

You guys are really helpful!

**skeeter** · September 26th 2009, 05:52 AM

Originally Posted by differentiate

Could someone check these for me? i was told to put g = 9.8 m/s for all calculus questions, and I think it's a bit ridiculous.

1. A long jumper running at 8 m/s jumps into the air, rising to a height of 0.7 metres. What is the length of the jump.

maximum height
$0.7 = \frac{V^2sin^2\alpha }{2g}$

$0.7 = \frac{8^2sin^2\alpha }{2\cdot 9.8}$

$13.72 = 64sin^2\alpha$

$sin\alpha = 0.463006479$

$\alpha = 27^{\circ}{35}'$

Range
$\frac{V^2sin2\alpha }{9.8}$

$=\frac{ 64sin55^{\circ}{10}' }{9.8}$

$= 5.36 m$

This is wrong. the answer says 5.29 m.

2. The velocity of a point moving along the x axis is given by
$V^2 = 10 + 8x - 2x^2$
Show that it is simple harmonic.

$\ddot{x} = \frac{d}{dx}(\frac{1}{2}v^2)$

$= \frac{d}{dx}(\frac{1}{2}10 + 8x - 2x^2)$

$= \frac{d}{dx}5+4x - x^2$

$= 4 - 2x$

However this is not the form $-n^2x$

so how would I do this?

Thanks!!!

You guys are really helpful!

(1) why so upset over a difference of 7 cm ? ... whoever came up with that "answer" used 10 for g in their calculations.

(2) $v^2 = 10+8x-2x^2$

$2v \frac{dv}{dt} = (8-4x)\frac{dx}{dt}$

$a = 4-2x = -(\sqrt{2})^2 \cdot (x-2)$

the equilibrium position is $x = 2$ instead of the "usual" $x = 0$

the motion is simple harmonic

**differentiate** · September 26th 2009, 04:28 PM

hey thanks skeeter! thats awesome

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