Question

A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 4.7 cm from its equilibrium position?

A)The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.
Select all that apply.
1. force constant k
2. total energy E
3. mass m
4. maximum velocity vmax
5. amplitude A
6. potential energy U at x
7. kinetic energy K at x
8. position x from equilibrium

B)What is the kinetic energy of the object on the spring when the spring is compressed 4.7 cm from its equilibrium position?

C)What is the potential energy U of the toy when the spring is compressed 4.7 cm from its equilibrium position?

Answer 1

Hi there!

Part A:

The only quantities explicitly given to us are:

3. mass (m)

4. Maximum velocity (vmax)

5. Amplitude (A)

8. Position x from equilibrium

Part B:

To solve, we must begin by calculating the force constant, 'k'.

We can use the following relationship:

$v = \sqrt{\frac{k}{m}(A^2-x^2)$

We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:

$\frac{v^2}{(A^2-x^2)} = \frac{k}{m}$

$\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44$

Now, we can find the velocity at 4.7cm (0.047m) using the equation:

$v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s$

Plug this value into the equation for kinetic energy:

$KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}$

Part C:

The potential energy of a spring is given as:

$U = \frac{1}{2}kx^2$

Find 'k' using the derived quantity above:

$\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m$

Now, calculate potential energy:

$U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}$