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Simple Harmonic Frequency Problem

A 2500kg steel sphere is suspended by a steel cable. The distance of suspension to the centre of mass of the sphere is 15m. The sphere is pulled a distance of 3m to one side and then released. The suspension cable is vertical when the sphere hits the wall. Calculate the speed of the sphere when it hits the wall. Assume g to be 10 ms-2

I'm really stuck on this problem, if anyone can help please do.

T = 2π * √l / g
T = 2π * √15 / 10
T = 7.6

f = 1 / T
f = 0.13Hz

Therefore...

v = -ω A sin ωt

A = 15m
ωt = 0.82 * 7.6 = 6.2
ω = 2πf = 0.82

v = -0.82 * 15 * sin 6.2
v = 0.62ms-1

The correct answer should be 2.45ms-1 and I have no idea how to get that. I was thinking that possibly my ω value is incorrect, and I should solve for ω using:

ωav = change in angular displacement (rad) / time taken (sec)

...but I can't seem to find out to calculate the change in angular displacement with the information I have.

Help please... :confused:
T = 2π * √l / g
T = 2π * √15 / 10
T = 7.6s

since it completes one-fourth of its oscillation

t=7.7/4=1.9s

Using
v = -ω A sin ωt
v = -(0.82)(3) sin(0.82x1.9)
v = 2.45m/s
Reply 2
Avatar for Iscariot
Iscariot
OP
11
Ok thanks a lot, but two questions:

a) - Why did you use 3m as the radius and not 15m?

b) - How do you know that 3m is one-forth of it's oscillation?

Thanks.
Reply 3
Avatar for Iscariot
Iscariot
OP
11
I realised A is equal to the displacement from the equilibrium, so yes A = 3m, but that still doesn't explain my other question.
To answer your second question, one oscillation is from max amplitude to max amplitude and back. the ball only travels one quarter of that distance
Reply 5
Avatar for Iscariot
Iscariot
OP
11
Thanks a lot!
Reply 6
Avatar for CossLne
CossLne
16
I'm doing Advanced Higher Physics and am considering doing an investigation on simple harmonic motion. I need 3 different experiments that prove 1 thing. HELP

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