Question

Week Seven Coursework 1 Classical Mechanics The figure shows a compound pendulum. A mass 3m is suspended from a fixed point, by a light, inextensible string of length L. A second mass m is suspended from the first by an identical string 1.1. For the system in its equilibrium position, as shown in the left-hand image, write down the tension in the two strings. [4 marks) 1.2. For small displacements, we may assume the tension in each string remains the same as at equilib- rium. With this assumption, find the horizontal force on each mass in terms of the coordinates : and y, indicated in the right-hand image. The coordinates are the horizontal displacements of the masses from their respective equilibrium positions. Small-angle approximations should be used. [7 marks] 1.3. Hence find the normal modes of the system and their corresponding frequencies. [9 marks)

Answer 1

1.1 Tension in the lower string T_l = mg

Tension on the upper string T_u = 3mg + mg = 4 mg

1.2. Horizontal force on the upper mass = T_u *Cos($\theta$) -  $\theta$ is the angle it makes with the horizontal

= 4mg*x/l

The lower string makes an angle (y-x)/l with the horizontal

Horizontal force on the lower mass = mg*(y-x)/l

1.3 Torque on the upper mass = -l* 4mg*x/l = -4 mgx

MI of the mass about the suspension I = ml²

Torque = I $\alpha$

$\alpha$ = -4 mgx /I = -4gx/l²  

for small angle we can approx. x= l$\theta$ ,

$\alpha$ = -4g/l  $\theta$

This is SHM with $\omega$ ² = 4g/l

angular frequency  $\omega$ = 2(g/l)^1/2  

for lower mass torque = -2l * mg*(y-x)/l = -2mg(y-x)

= I$\alpha$

for the lower mass I = m(2l)² = 4ml²  

$\alpha$ = -2mg(y-x) /4ml² ; for small angles (y-x) = l$\theta$

=- ( g/2l)$\theta$  

= -$\omega$²$\theta$ ( SHM)

angular frequency $\omega$ = (g/2l)^1/2  

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