1998 hsc shm (1 Viewer)

[Smilebuffalo]

3.c) A particle moves in a straight line and its position at time t is given by:

x=1+sin4t+ (root3)cos4t

(i) Prove that the particle is undergoing simple harmonic motion about
x = 1.


When i was reading the exam report, they said it was incorrect to do the following:

a = −16 sin 4t − 16(root3)cos 4t
= −16(sin 4t + (root3)cos4t)
= −16(x − 1) .

This got me confused. I assumed we were always meant to show a = -n^2(x-b)

can someone explain?

 

[jet]

If you look at it, they haven't factorised the negative out of the term in cos 4t in the marking notes. They're just pointing out a mistake that many candidates made.

That method is perfectly fine, though I would suggest adding in an extra step just so that you don't confuse a) the markers and b) you.



So that you can prove to the markers that you knew what you were doing and weren't just 'fudging' the result.

 

[Schoey93]

If you look at it, they haven't factorised the negative out of the term in cos 4t in the marking centre. They're just pointing out a mistake that many candidates made.

That method is perfectly fine, though I would suggest adding in an extra step just so that you don't confuse a) the markers and b) you.



So that you can prove to the markers that you knew what you were doing and weren't just 'fudging' the result.

Jetblack, you're Josh V ernon's friend; Ashkon, aren't you??

 

[jet]

Jetblack, you're Josh V ernon's friend; Ashkon, aren't you??

That would be a no. The names Michael mate.

 

[tku336]

Maybe they want you to be really obvious at put the 16 as 4^2?

 

[tku336]

Or explain why once its in that form it's SHM?

 

[Trebla]

3.c) A particle moves in a straight line and its position at time t is given by:

x=1+sin4t+ (root3)cos4t

(i) Prove that the particle is undergoing simple harmonic motion about
x = 1.


When i was reading the exam report, they said it was incorrect to do the following:

a = −16 sin 4t − 16(root3)cos 4t
= −16(sin 4t + (root3)cos4t)
= −16(x − 1) .

This got me confused. I assumed we were always meant to show a = -n^2(x-b)

can someone explain?

Read it a bit more carefully. It actually says this was a common error:
a = −16 sin 4t − 16(root3)cos 4t
= −16(sin 4t − (root3)cos4t)
= −16(x − 1)

 

[Smilebuffalo]

oooh i see now. I was reading the exam report and marking a past exam late at night so i must've missed the minus sign. Thank you very much for pointing that out :uhhuh: