SHM - Questions (1 Viewer)

[currysauce]

1. The period of a particle moving in SHM is 6s and its amp. is 8cm. Calculates its velocity and acceleration (to 1.d.p) when the displacement is 5cm from the centre of its motion.

ans but i can't show:

vel:-6.5
accl: -5.5

2. A particle moves in a straight line so that its accl. at any time is d²x/dt² = -9x. Find its persion, amp. and displacement at time t if initially the particle is 2cm from the orogin and has velocity 2root(3) cm/s

ans: p: 2pi/3
amp: 4root(3)/3

x= 4root(3) cos (3t- pi/6)

3. A particle moves in SHM with amp. 5cm and period 6secs. Find

a) the velocity when the particle is 2.5cm from the centre of the motion
b) the maximum acceleration

a ans: - 5root(3)pi / 6

b ans: 5pi²/9



thanks all

 

[Jumbo Cactuar]

Q1

x = n sin ((2.pi/w).t)
period (w) = 6 secs
magnitude (n) = 8 cm
x = 8 sin ((2.pi/6).t)
v = dx/dt = (16.pi/6) cos ((2.pi/6).t)
a = dv/dt = -(32.pi²/36) sin ((2.pi/6).t)

when x = 5cm;
t = (6/2.pi) sin^-1(5/8)
v = (16.pi/6) cos sin^-1(5/8) = 6.5cm/sec (also -6.5)
a = -(32.pi²/36) * (5/8) = -5.5cm/sec.sec (also 5.5)

 

[shafqat]

SHM

At extremeties,
x = amplitude
v = 0
a = max

at origin,
x = 0
v = max
a = 0

For the form a = -n^2 x, let a = d/dx(v^2/2)
integrate for v, and use any information they give and the table above to find C

if a particle moves about a point p other than the origin, replace x by (x - p)

Example: to find amplitude, integrate to find the relation between v and x, then put v = 0

Trig Form

x= a sin nt, where n = 2pi/T, T = period
use -sin, cos, if the particle doesnt start and origin

Note: results such as v^2 = n^2(a^2 - x^2) should not be assumed in a question, the only equations you can assume are the original definition of SHM, that is a is proportional to x, and the trig form

 

[currysauce]

and so the last 2 remaining questions
?

 

[currysauce]

anyone!@ k

 

[shafqat]

2. A particle moves in a straight line so that its accl. at any time is d²x/dt² = -9x. Find its persion, amp. and displacement at time t if initially the particle is 2cm from the orogin and has velocity 2root(3) cm/s

ans: p: 2pi/3
amp: 4root(3)/3

x= 4root(3) cos (3t- pi/6)

d/dx(v^2/2) = -9x
v^2/2 = -9x^2/2 + C/2
v^2 = -9x^2 + C
when x = 2, v = 2root3
so 12 = -36 + C
C = 48
v^2 = -9x^2 + 48
put v = 0, x^2 = 48/9 = 16/3
x = 4/root3 = 4root(3)/3
period = 2pi/n = 2pi/3, as n^2 = 9 from the original equation

Exactly as i exaplained is the post above.

"Give a man a fish and you feed him for a day. Teach a man to fish and you feed him for a lifetime."

 

[tempco]

i thought you were gonna say "give em a finger and they take an arm".

oh well.