Question

The cart is originally pulled by distance x0 (maximum contraction of the ventricular wall) from equilibrium and released. If the valve opened when the contraction of the muscle was half way between the maximum stretch and the equilibrium position, what is the speed, v , of the cart (blood flow speed) at that instance? This is a symbolic question. Express your results in terms of the amplitude x0 , spring constant k , and the mass of the cart m .

HINTS: What type(s) of energy is/are present when the cart is released? What types(s) of energy is/are present when the cart is halfway to equilibrium? When you have your final expression, perform a "sanity check": will the velocity at this point be greater, less than, or equal to the velocity at equilibrium or the point of maximum contraction? Does your symbolic answer reflect this fact?

Answer 1

From the given information, it would be modeled as the spring mass system

let the spring of original length L , is fixed at one end and other end is attached with a mass(CART)

when a force applied so that it stretches to X0 , then Elastic potential energy will be developed in the spring.

when the applied force by cart is removed, this energy tries the spring to come to its equilibrium position, acts in opposite direction to the applied force.

when it reaches the equilibrium position the complete elastic potential energy converts in to kinetic energy, so that at mean position , the cart have maximum velocity .

when it is half the way to equilibrium position the total energy of the system is sum of kinetic energy and elastic potential energy

E_half way = E.P.E +k.e

E_equilibrium = k.e

we know that the E.P.E = 0.5*k*X0^2 , k is spring constant , X0 is stretch in the spring

k.e = 0.5*m*v^2