• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

SHM and projectile problems- help needed (1 Viewer)

xxmissy

New Member
Joined
Oct 22, 2006
Messages
28
Gender
Female
HSC
2007
Hey guys I need help with these:confused: ... or how i would approach it:

Simple harmonic motion:

The period of a particle in moving in SHM in 6 s and its amplitude is 8cm. Calculate its velocity and acceleration (correct to 1 decimal place) when the displacement is 5cm from the centre of the motion.


Projectiles:

A horizontal drainpipe 6m above sea level empties stormwater into the sea. It the water comes out horizontally and reaches the sea 2m out from the pipe, find the initial velocity of the water, correct to 1 decimal place. Let g be 10m/s^2 and neglect air resistence.

Help would be very much appreciated!!
 
Joined
Mar 3, 2005
Messages
2,359
Location
Wollongong
Gender
Male
HSC
2006
xxmissy said:
Projectiles:

A horizontal drainpipe 6m above sea level empties stormwater into the sea. It the water comes out horizontally and reaches the sea 2m out from the pipe, find the initial velocity of the water, correct to 1 decimal place. Let g be 10m/s^2 and neglect air resistence.

Help would be very much appreciated!!
i cant remember the variables they use in maths ext 1 or the methods used but i can tell you the physics method...hopefully you can follow...

let u = initial, v = final velocity, y = displacement in vertical direction, x = displacement in vertical direction, a = acceleration, t = time.

ok you need to find time.

consider the y direction first.

y = ut + 0.5at^2

u = 0, so y = 0.5at^2

solve for t, uve got height and acceleration due to gravity so plug them in and you got t.

t = sqrt(2y/a)

now consider the x direction.

you should know that

x = ut + 0.5at^2 ----> x = ut, since a = 0 in the x direction

uve got x, youve got t, so solve for u and thats ur intial velocity

sorry its kinda cluttered, hope you followed that.
 
Last edited:

elseany

Member
Joined
Aug 2, 2005
Messages
83
Location
Eastern Suburbs
Gender
Female
HSC
2007
For the first question:

Period:
2(pi)/n = 6
n = (pi)/3
a = 8

we can use the following formula; v2 = n2(a2 - x2)
substitute in our values for n and a and let x = 5

we get v2 = 13(pi)2/2
so v = 6.5cm/s

for the second part you use the formula; x'' = -n2x
so just again sub in our value for n and x=5
and u should get x'' = a = -5(pi)2/9
so a = -5.5cm/s/s
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top