S.h.m (1 Viewer)

[shinji]

i just can't get the amplitude of the motion .. all except for that is great for it..

here's a lil q,
X = 5 + 8sin2t + 6cos2t

i proved that it undergoes SHM
where a= -4(x-5)

but i don't knw how to find the amplitude
help ne1?

 

[Trev]

Hint:
asinx+bcosx=sqrt(a²+b²).sin[x+tan^-1(b/a)].

 

[risaka]

u tried using the tranformation method??
x-5 = 8sin2t + 6cos2t
= asin(2t + z)
= asin2tcosz + acos2tsinz

after a bit of calculation...should turn out something like:
x - 5= 10sin(2t+36degrees 52')
therefore amplitude = 10??

 

[sasquatch]

The amplitude is 15.

You can work it out as such:

x = 5 + 8sin2t + 6cos2t

the angles for both the sine and cosine functions are the same, so the maximum value for this part of the statement "8sin2t + 6cos2t" is given as A = root(a^2 + b^2) as trev stated.

So you add that 10 to the 5, so the amplitude is equal to 15.

Oh yeah, you can do it risaka's way.. looks better, but he didnt do this:

x - 5= 10sin(2t+36degrees 52')
x = 5 + 10sin(2t+36degrees 52')

maximum value of sin(2t+36degrees 52') = 1, so at maximum value x = 5 + 10 = 15.

hence amplitude is 15.

 

[Riviet]

Remember that the amplitude is not the maximum displacement from the origin but the maximum displacement from the centre of oscillation.
If a particle's displacement is expressed as a function of time, such as in the form x = B + Asin(nt + @)

A is the amplitude. If B>0, the particle's centre of oscillation is shifted in the positive direction of displacement by B units. If B<0, the particle's centre of oscillation is shifted in the negative direction of displacment by B units.

Therefore, the amplitude of a particle whose motion is defined by x = 5 + 10sin(2t+36degrees 52') has an amplitude of 10 units.

Note that B does NOT affect the amplitude of the particle's motion.

Hope that clears things up.