a) To find the Euler-Lagrange equations for this system, we need to set up the Lagrangian. Let’s denote the position of the mass M as x and the position of the mass μ as y. The Lagrangian can be written as:

L = T – V

where T is the kinetic energy and V is the potential energy. The kinetic energy of the system can be written as:

T = 1/2 M (dx/dt)^2 + 1/2 μ (dy/dt)^2

where dx/dt and dy/dt are the velocities of the masses. The potential energy can be written as:

V = Mgy + μgy – T

where g is the acceleration due to gravity and T is the tension in the string. Note that the potential energy includes both the gravitational potential energy of the masses and the potential energy stored in the string due to its extension.

Using the Law of Cosines, we can relate the position of the masses to the length of the string, l:

l^2 = x^2 + y^2 – 2xy cos(30°)

Differentiating this equation with respect to time, we get:

2l dl/dt = 2x dx/dt + 2y dy/dt cos(30°)

Substituting y = l – d and solving for x, we get:

x = (l^2 – (l-d)^2 – q^2)/(2q)

Substituting this into the Lagrangian, we get:

L = 1/2 M ((l^2 – (l-d)^2 – q^2)/(2ql))^2 + 1/2 μ (dl/dt)^2 – Mg(l-d) – μg(l-d) + T

Now we can use the Euler-Lagrange equations:

d/dt (dL/d(dl/dt)) – dL/dl = 0

to find the equations of motion for the system. Taking the derivative of L with respect to dl/dt, we get:

dL/d(dl/dt) = μ dl/dt

Taking the derivative of L with respect to l, we get:

dL/dl = -Mg – μg + (M(dl/dt)^2 – (l^2 – (l-d)^2 – q^2)/(2ql)^3 M dl/dt^2)

Plugging these into the Euler-Lagrange equations, we get:

μ d^2y/dt^2 = μg – T

M d^2x/dt^2 = Mg – T – M ((x/l)^2 – ((l-d)/l)^2 – (q/l)^2)/(2q) x/l^3

b) To find the position of the mass M as a function of time, we need to solve the equation of motion for x:

M d^2x/dt^2 = Mg – T – M ((x/l)^2 – ((l-d)/l)^2 – (q/l)^2)/(2q) x/l^3

We can simplify this equation by defining the variable u = x/l, and noting that l = sqrt(x^2 + (l-d)^2 – 2x(l-d)cos(30°)):

M d^2u/dt^2 = Mg – T – M ((u^2 – ((l-d)/l)^2 – (q/l)^2)/(2q)) u/l^3

This is a second-order differential equation for u, which can be solved numerically or with appropriate approximations.

（c）To find the tension in the string, we can use the equation of motion derived in part (b) and solve for the tension at any given point in time.

From part (b), we found that the acceleration of the system is given by:

a = g(sinθ – μcosθ)

At any given point in time, the tension in the string is equal to the force required to accelerate the mass M at the same rate as the acceleration of the system. Therefore, we have:

T = Ma

Substituting the expression for acceleration obtained above, we get:

T = M[g(sinθ – μcosθ)]

Thus, the tioension in the string is dependent on the angle θ and the mass rat μ/M.