A0.42 kg mass is attached to a light spring with a force constant of 34.9 n/m and set into oscillation on a horizontal frictionless surface. if the spring is stretched 5.0 cm and released from rest, determine the following. (a) maximum speed of the oscillating mass .45578 correct: your answer is correct. m/s (b) speed of the oscillating mass when the spring is compressed 1.5 cm .43478 correct: your answer is correct. m/s (c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position .43478 correct: your answer is correct. m/s (d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value

(a) 0.456 m/s

The maximum speed of the oscillating mass can be found by using the conservation of energy. In fact:

- At the point of maximum displacement, the mechanical energy of the system is just elastic potential energy:

(1)

where

k = 34.9 N/m is the spring constant

A = 5.0 cm = 0.05 m is the amplitude of the oscillation

- At the point of equilibrium, the displacement is zero, so all the mechanical energy of the system is just kinetic energy:

(2)

where

m = 0.42 kg is the mass

vmax is the maximum speed, which is maximum when the mass passes the equilibrium position

Since the mechanical energy is conserved, we can write (1) = (2):

(b) 0.437 m/s

When the spring is compressed by x = 1.5 cm = 0.015 m, the equation for the conservation of energy becomes:

(3)

where the total mechanical energy can be calculated at the point where the displacement is maximum (x = A = 0.05 m):

So, solving (3) for v, we find the speed when x=1.5 cm:

(c) 0.437 m/s

This part of the problem is exactly identical to part b), since the displacement of the mass is still

x = 1.5 cm = 0.015 m

So, the speed when this is the displacement is

(d) 4.4 cm

In this case, we have that the speed of the mass is 1/2 of the maximum value, so:

And by using the conservation of energy again, we can find the corresponding value of the displacement x:

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