Eliminate T to Show SHM of a Rectangular Plate

Avi1995
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Q:A rectangular plate of sides a and b is supended from a ceiling by two parallel strings of length L each(Fig). The Separation between the strings is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute SHM. Find the time period

http://www.freeimagehosting.net/newuploads/z8fe5.jpg
Basic SHM equations:
1.F=-kx
2.T=(2*pi)/(omega)
Let the plate be displaced a slightly for a displacement x and the angle of strings with vertical be theta.
2Tsin(theta)=-ma
Using Approx.
sin(theta)=x/L
-2Tx/L=-ma
I don't know how to eliminate T from the equation. Plz help!
 
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I don't know how to eliminate T from the equation. Plz help!
How would you do this for a normal pendulum?
Have you sketched the situation when the block has a small displacement?
Which direction should the unbalanced force act?
 
Thank you sir! I finally solved it!
Solution:

After I sketched it I found what I was doing wrong.
Taking forces normal to string,
-mgsin(theta)=ma
a=-g(theta)
theta=x/L(approx)
a=-g/L
T=2*pi*sqrt(L/g)
 
Drawing pictures is extremely useful that way ;)
Always draw the picture.

aside: how does this result compare with a simple pendulum?
 
It is exactly similar!
 
It's exactly similar eh?
One of the useful things you can do with these exercises is see how the result compares with other things you know about. In homework, actually writing these observations down, briefly, under the answer will often get you extra marks ;)

It's also a useful reality-check - does it make sense that the period of this sort of pendulum is exactly the same form as for a simple pendulum?
At some point you will be expected to deal with problems that nobody has solved before.
This sort of discipline is one way to be confident you got the right results.
 
Simon Bridge said:
It's exactly similar eh?
One of the useful things you can do with these exercises is see how the result compares with other things you know about. In homework, actually writing these observations down, briefly, under the answer will often get you extra marks ;)

It's also a useful reality-check - does it make sense that the period of this sort of pendulum is exactly the same form as for a simple pendulum?
At some point you will be expected to deal with problems that nobody has solved before.
This sort of discipline is one way to be confident you got the right results.
Err...I mean it was exactly same :P. In my mind it does make little sense, as the mass is symmetrically suspended. Since time period of pendulum was independent of mass this should be also.
 
Hmm... English a second language?
Compare:
In my mind it does make little sense
... with
In my mind it does make a little sense​

You can usually get away with being really sloppy with grammar - but there are a few pitfalls. Compare: "I have little skill with other languages" with "I have a little skill with other languages."

If you hadn't elaborated, a reader would certainly have taken the opposite meaning to what you intended.

Back to physics though:
If you look at the motion of the center of mass of the block you'll see it is an arc, just like the motion of the ideal pendulum bob. You could also have done the problem by resolving the tensions into components through the center of mass and perpendicular to that line (using a free-body diagram).

Have fun :)
 

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