Applied Maths - Mechanics 46

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Let A be the acceleration of pulley.
Let Ar be the relative acceleration of m and 2m respect to the pulley.
Let T be the tension in the string.
Consider 2m,
2mg - T = 2m(Ar - A)
Ar = [(2mg - T)/(2m) + A]...............(1)
Consider m,
T - mg = m(A + Ar)...................(2)
Sub (1) into (2),
T - mg = m[(2mg - T)/(2m) + A + A]
T = 4m(g+A)/3...........................(3)
Consider the pulley,
-kx - 2T - mg = mA, where k = 12mg/a
-12mgx/a - 2T - mg = mA.................(4)
Sub (3) into (4),
-12mgx/a - 2[4m(g+A)/3] - mg = mA
A = - 36gx/(11a) - g
Changing the variable x with h, where h is the displacement from the equilibrium position.
We have,
A = - 36gh/(11a)
So, A is proportional to h.
i.e. The motion is S.H.M.
Compare A= - ω2 h
ω = √[36g/(11a)]
Period = 2(pi)/ω = 2(pi)√[11a/(36g))] = [(pi)/3]√(11a/g)

2009-08-16 10:41:14 補充：
The acceleration of the pulley is different from your ans. I don't whether I am correct

2009-08-16 11:05:07 補充：
I think I was wrong before. here comes a correct version:

http://www.flickr.com/photos/27778998@N04/3825265506/

2009-08-16 11:05:17 補充：
Let A be the acceleration of pulley.
Let Ar be the relative acceleration of m and 2m respect to the pulley.
Let T be the tension in the string.

Consider 2m,
2mg - T = 2m(Ar + A)....................(1)

Consider m,
mg - T = m(A -Ar)...................(2)

2009-08-16 11:05:25 補充：
Solving (1) and (2),
we have T = 4m(g - A)/3.............(3)

Consider the pulley,
-kx + 2T + mg = mA, where k = 12mg/a
-12mgx/a + 2T + mg = mA.................(4)

2009-08-16 11:05:34 補充：
Sub (3) into (4),
-12mgx/a + 2[4m(g - A)/3] + mg = mA
A = g - 36gx/(11a)

Changing the variable x with h, where h is the displacement from the equilibrium position.
We have,
A = - 36gh/(11a)
So, A is proportional to h.
i.e. The motion is S.H.M.

2009-08-16 11:05:42 補充：
Compare A= - ω2 h

ω = √[36g/(11a)]

Period = 2(pi)/ω = 2(pi)√[11a/(36g))] = [(pi)/3]√(11a/g)

2009-08-16 11:32:33 補充：
Ignore the old version, it is WRONG.

No diagram for this question .

I do not understand the question, is a diagram available?

2009-08-16 11:39:05 補充：
Taking the downward direction as positive.
Let the position of mass m below the hanging point be y
Let the position of mass 2m below the hanging point be z
Let the tension in the inelastic string be T

2009-08-16 11:39:45 補充：
Length of the string = y – (a + x) + z (a + x) is a constant
Therefore d2y/dt2 + d2z/dt2 = 2d2x/dt2 … (1)
The tension in the spring is -12mgx/a

2009-08-16 11:40:00 補充：
Force equations of the pulley, m and 2m respectively are:
-12mgx/a + 2T + mg = md2x/dt2 … (2)
mg – T = md2y/dt2 … (3)
2mg – T = 2nd2z/dt2 … (4)

2009-08-16 11:40:13 補充：
(3) => d2y/dt2 = g – T/m … (5)
(4) => d2z/dt2 = g – T/2m … (6)
Combining (1), (5) and (6) => g – T/m + g – T/2m = 2d2x/dt2
2mg – 2T + 2mg – T = 4md2x/dt2
T = 4mg/3 – (4m/3)d2x/dt2 … (7)

2009-08-16 11:40:23 補充：
Sub (7) into (2),
-12mgx/a + 2[4mg/3 – (4m/3)d2x/dt2] + mg = md2x/dt2
-12gx/a + 8g/3 – (8/3)d2x/dt2 + g = d2x/dt2
-12gx/a + 11g/3 = (11/3)d2x/dt2
d2x/dt2 = g – 36gx/11a => SHM
Period = 2π√(11a/36g) = (π/3)√(11a/g)