Question

A partridge of mass 5.08 kg is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.23 s .

1.What is its speed as it passes through the equilibrium position?

2.What is its acceleration when it is 0.050 m above the equilibrium position?

3.When it is moving upward, how much time is required for it to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it?

4.The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

Answer 1

Solution

1. To solve this problem is neccesary the equation of the particle

In this case, the speed of the particle in the equilibrium position is maximum

Therefore

In the maximum point

We calculate the angular frequency

Therefore

2.  Now, the equation of the partice in the x=0.050m above the equilibrium position is obtained, as follows

for x=0.05m

Now, for x=-0.05m

3.

and

reorganizing

4.

.

Clearing k

We use the Hooke Law