question from catholic trail 05 (1 Viewer)

[damo676767]

7 a i

does any one have that question/ the answer

any help would be grealty appreaciated

 

[evette13]

Yes, the projectile one?

 

[evette13]

q7ai is the easy part...do u mean 7b?

 

[damo676767]

the simple harmonic one

at some point we had to convert an equation with both sin and cos in it, in to just cos

 

[evette13]

Thats question 6 a:
"a partivle is performing SHM about a fixed point O on a straight line. At time t seconds it has displacement x metres from O gien by x= cos2x-sin2t
i) Express x in the form Rcos(2t+@) for som R>0 and 0<@<pi/2
ii) find the amplitude and period
iii) determine whether the particle is initially moving towards O or away from O and whether it is initially speeding up or slowing down
iv) find the time at which the particle first returns to its starting point

 

[damo676767]

x= cos2x-sin2t

thanyou very much

but is this corect, or is it x = cost 2t - sin 2t

 

[evette13]

Answer?

Spoiler

Rcos(2t+@) ....therefore R=sqroot (1^2 +1^2) = sqroot 2
Tan @ = -1/1=-1 therefore @ = 3pi/4, 7pi/4 (etc)
For @ btw 0 and pi on 2, @ must equal pi/4
Thus, substituted in Rcos (2t-@)
Therefore, x=sqroot2 cos (2t+pi/4)

 

[evette13]

Sorry it is cos2t-sin2t

 

[damo676767]

thats cool, i just didn't want to make the problem ia hold lot harder, if it was cos2x then that would be insane

 

[evette13]

oh by the way that is wat i wrote and got 2 marks, the catholic answers said:

x= cos2t-Sin2t
x= sqroot2 (1/sqroot2 cos2t - 1/sqroot2 sin2t)
= sqroot2 (cos pi/4*cos2t - sin pi/4*sin2t)
=srroot2 cos(2t+pi/4)

 

[damo676767]

oh by the way that is wat i wrote and got 2 marks, the catholic answers said:

x= cos2t-Sin2t
x= sqroot2 (1/sqroot2 cos2t - 1/sqroot2 sin2t)
= sqroot2 (cos pi/4*cos2t - sin pi/4*sin2t)
=srroot2 cos(2t+pi/4)

could you please explain that, i have no idea what your doing there

thankyou

 

[Bellow]

could you please explain that, i have no idea what your doing there

thankyou

x=cos2t - sin2t
= √2/√2 (cos2t - sin2t) (multiply top and bottom by √2)
= √2 (1/√2 x cos2t - 1/√2 x sin2t)
= √2 (cos(pi/4)xcos2t - sin(pi/4)xsin2t) since cos(pi/4)=1/√2 and sin(pi/4)=1/√2
= √2 (cos(2t+pi/4)) using the trig identity cos(a+b)=cosaxcosb-sinaxsinb
= √2 cos(2t+pi/4)

 

[damo676767]

ingenious

thakyou all very much