SYDNEY BOYS - Multiple Choice Questions (1 Viewer)

[hscwav2012]

SBHS 2012 - Q4,5,7

Need help with Q 4 , 5 & 7.

https://docs.google.com/file/d/0B4HdQ56p9QhLQW9RQ1BkaVF5R00/edit?pli=1

Here's Q7 - I couldn't copy Q4 or 5 so attached them in the link.

7. A particle moving in Simple Harmonic Motion oscillates about a fixed point O in a straight line
with a period of 10 seconds. The maximum displacement of P from O is 5 m. Which of the
following statements is/are true?

If P is at O moving to the right, then 22 seconds later P will be:

(I) moving towards O.

(II) moving with a decreasing speed

(III) at a distance 5sin(22pi/5) m to the right of O

(A) - (I), (II) and (III)

(B) - (I) and (II) only.

(C) -(II) and (III) only.

(D) - none of the above.

 

[ahdil33]

Okay.

For Q4. tan(pi/2 +x) = -tan(pi/2-x)=-cot(x)

as sin x = k , now we can draw a little triangle where the angle is x, the opposite side is k, and the hypotenuse is 1. this means the adjacent side is root(1-k^2).

therefore, cos(x) = root(1-k^2)

as -cot(x) = -cos(x)/sin(x)= -root(1-k^2)/k, which is C.

--

Q5. Still don't understand this one lol, guessed it right though.

Q7. We're dealing with SHM over here, let's see if we can get enough details to make an equation. So T = 10 seconds

T = 2pi/n, therefore n = pi/5.

ampltitude, a = 5m

For I), we know that it begins moving to the right at 0, when t= 0. it has a turning point at 2.5, 5, 7.5, then come back to the centre at t=10. this happens again, until t= 20, another period goes by. at t=22, it would still be moving right before the first turning point in the cycle. TRUE

II )At t=22, it's slowing down to reach the turning point at t=22.5, so yeah the speed must be decreasing. TRUE

for III) I'd make an equation for this, since it's starting at O, x = 5sin((pi/5)t) would be the equation. so yeah, this is true as well, to the right of O.

so A.

 

[hscwav2012]

the solution seem to have other solutions https://docs.google.com/file/d/0B4HdQ56p9QhLZjRtWDZYa1lYR2M/edit?pli=1

4a
5a
7c