A block of mass M hangs from a rubber cord. The block is supported so that the cord is not stretched. The unstretched length of the cord is L0 and its mass is m, much less than M. The "spring constant" for the cord is k. The block is released and stops at the lowest point. (Use L_0 for L0, M, g, and k as necessary.)

(a) Determine the tension in the cord when the block is at this lowest point.
1

(b) What is the length of the cord in this "stretched" position?
2

(c) Find the speed of a transverse wave in the cord, if the block is held in this lowest position.

(a) At the lowest point (if motionless), the cord tension equals the weight, M g.

(b) The stretched lendth of the cord is L = L0[1 + (Mg/k)] , since it stretches by an amount L0*(Weight)/k

(c) V = sqrt[Mg/(m/L)] = sqrt(LMg/m)

this is wrong

What is the right answer? Do you know?

T=2*M*g

I can't get the other parts.

can you get the answers to b and c?

To solve this problem, we can use the concepts of equilibrium and the properties of a spring to find the answers.

(a) Determining the tension in the cord when the block is at the lowest point:
When the block is at the lowest point, the tension in the cord must balance the weight of the block. We can set up an equation to represent this equilibrium:
Tension in the cord = Weight of the block

The weight of the block can be calculated using the equation:
Weight = mass × gravity

In this case, the mass of the block is M and the acceleration due to gravity is g. So, the weight of the block is M × g.

Therefore, the tension in the cord is equal to M × g.

(b) Finding the length of the cord in the stretched position:
When the block is at the lowest point, the rubber cord is stretched. The length of the cord in the stretched position can be determined by considering the equilibrium of forces acting on the cord.

At the lowest point, the elongation of the cord can be represented as ΔL = L - L0, where L0 is the unstretched length and L is the stretched length of the cord.

By Hooke's Law, the force exerted by a spring is given by:
Force = spring constant × elongation

In this case, the spring constant is k and the elongation is ΔL.

Therefore, the force exerted by the cord is k × ΔL.

Since the tension in the cord is equal to the force exerted by the cord, we can equate the two and solve for the stretched length:
Tension in the cord = k × ΔL

From part (a), we know that the tension in the cord is M × g.

Therefore, M × g = k × ΔL.

Solving for ΔL will give us the elongation of the cord. Adding this elongation to the unstretched length will give us the stretched length of the cord.

(c) Finding the speed of a transverse wave in the cord, when the block is held in the lowest position:
To find the speed of a transverse wave in the cord, we can use the equation:
Wave speed = √(tension / linear mass density)

The tension in the cord can be obtained from part (a) as M × g.

The linear mass density of the cord is given by:
Linear mass density = mass per unit length = m / L0

Therefore, the wave speed is given by:
Speed = √(M × g / (m / L0))

By simplifying the expression, we get:
Speed = √(M × g × L0 / m)

So, by using the values of M, g, L0, and m, we can calculate the speed of the transverse wave in the cord when the block is held in the lowest position.