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[hscuserr]

A particle is moving in a straight line according to the equation
x = 5 + 6 cos 2t + 8 sin 2t,
where x is the displacement in metres and t is the time in seconds.
(i) Prove that the particle is moving in simple harmonic motion by showing 2
2 that x satisfies an equation of the form x

= −n ( x − c).
(ii) When is the displacement of the particle zero for the first time? 3


in questions (ii),

how do you Rewrite 6cos2t + 8sin2t = −5 as 10cos(2t − α) = −5
is there any reason why it becomes cos(2t - a)

 

[calamebe]

Well, you can expand 10cos(2t-α) to 10cos(2t)cos(α)+10sin(2t)sin(α), and then equate one of the terms. So for example, the coefficient of sin(2t) on both sides is 10sin(α)=8, so α=arcsin(4/5). You could also write this is α=arccos(3/5) or α=arctan(4/3), and if you want to see why you can draw a right angled triangle with sides 3, 4 and 5. The reason we put it in that cos(2t-α) or sin(2t+α) form is that it is so much easier to work with.

 

[InteGrand]

A particle is moving in a straight line according to the equation
x = 5 + 6 cos 2t + 8 sin 2t,
where x is the displacement in metres and t is the time in seconds.
(i) Prove that the particle is moving in simple harmonic motion by showing 2
2 that x satisfies an equation of the form x = −n ( x − c).
(ii) When is the displacement of the particle zero for the first time? 3


in questions (ii),

how do you Rewrite 6cos2t + 8sin2t = −5 as 10cos(2t − α) = −5
is there any reason why it becomes cos(2t - a)

,

 

[davidgoes4wce]

A particle is moving in a straight line according to the equation
x = 5 + 6 cos 2t + 8 sin 2t,
where x is the displacement in metres and t is the time in seconds.
(i) Prove that the particle is moving in simple harmonic motion by showing 2
2 that x satisfies an equation of the form x = −n ( x − c).
(ii) When is the displacement of the particle zero for the first time? 3


in questions (ii),

how do you Rewrite 6cos2t + 8sin2t = −5 as 10cos(2t − α) = −5
is there any reason why it becomes cos(2t - a)

Do you have an answer for it? Just want to compare my answers

 

[calamebe]

Do you have an answer for it? Just want to compare my answers

You'd get the general solution being : t = (3arcsin(4/5) + 2pi(3n±1))/6, and from that the first time is t = (3arcsin(4/5) + 2pi)/6, or roughly t = 1.51.

 

[Nexa579]

,

Are we (3U) allowed to just write and without showing the equating coefficients with the trig identities?

Unsure because pretty much all the worked solutions don't use the above way but I personally use it, so not sure if HSC will accept?

 

[calamebe]

Are we (3U) allowed to just write and without showing the equating coefficients with the trig identities?

Unsure because pretty much all the worked solutions don't use the above way but I personally use it, so not sure if HSC will accept?

That's what my teacher taught us, so I'm guessing it's allowed.

 

[InteGrand]

You also generally need to be careful with the arctangent.

 

[InteGrand]

• atan2 function: https://en.wikipedia.org/wiki/Atan2#Definition_and_computation .