Question

With the given information, it is impossible to choose between the positive and negative solutions in part (b). Notice that the sum KE + in part (c) equals the total energy E found in part (a), as it should (except for a small discrepancy due to rounding).

PES

QUESTION:

Does doubling the initial displacement double the speed of the object at the equilibrium point? Explain.

(a) Yes, the maximum speed is directly proportional to the initial displacement.

(b) No, it multiplies the speed at the equilibrium point by $\sqrt{2}$ .

(c) No, it multiplies the speed at the equilibrium point by 4.

(d) No, the speed at the equilibrium point is unaffected.

PRACTICE IT:

Use the worked example above to help you solve this problem. A 0.470 kg object connected to a light spring with a spring constant of 18.5 N/m oscillates on a frictionless horizontal surface.

(a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm.

E = _____ J

$V_{max}$ = _____ m/s

(b) What is the velocity of the object when the displacement is 2.00 cm?

_____ ± m/s

(c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm.

KE = _____ J

= _____ J

PES

EXERCISE:

For what values of x is the speed of the object 0.12 m/s?

x = ± _____ cm

Answer 1

DATA:

Solution:

Part a)

The total energy of the system is defined as:

${\color{Red} E=\frac{1}{2}kA^{2}}$

Replacing values given in DATA:

$E=\frac{1}{2}(18.5\, N/m).(0.0300\, m)^{2}$

Therefore we obtain:

${\color{Blue} E=8.33\times 10^{-3}\, J}$

While that, the maximum speed is defined as:

${\color{Red} v_{max}=\sqrt{\frac{k}{m}}.A}$

Replacing values we have:

$v_{max}=\sqrt{\frac{18.5\, N/m}{0.470\, kg}}.(0.0300\, m)$

Therefore we obtain:

${\color{Blue} v_{max}= \, 0.188\, m/s}$

Part b)

the velocity of the object is defined as:

${\color{Red} v_{x}=\pm \sqrt{\frac{k}{m}}.\sqrt{A^{2}-x^{2}}}$

Replacing values:

$v_{x}=\pm \sqrt{\frac{18.5\, N/m}{0.470\, kg}}.\sqrt{(0.0300\, m)^{2}-(0.0200\, m)^{2}}$

Thereore:

${\color{Blue} v_{x}=\pm\, \, 0.140\, m/s}$

Part c)

The Kinetic Energy is defined as:

${\color{Red} K=\frac{1}{2}m{v_{x}}^{2}}$

Replacing values:

$K=\frac{1}{2}(0.470\, kg)(0140\, m/s)^{2}$

Therefore:

${\color{Blue} K=4.61\times 10^{-3}\, J}$

While that, the potential energy is defined as:

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

Replacing values:

$U=\frac{1}{2}(18.5\, N/m).(0.0200\, m)^{2}$

Therefore we obtain:

${\color{Blue} U=3.70\times 10^{-3}\, J}$

Solution Exercise

The velocity is defined as:

${\color{Red} v_{x}=\pm \, \sqrt{\frac{k}{m}}.\sqrt{A^{2}-x^{2}}}$

isolating we have:

$x=\pm \, \sqrt{A^{2}-\frac{m{v_{x}}^{2}}{k}}$

Replacing values we have:

$x=\pm \, \sqrt{(0.0300\, m)^{2}-\frac{(0.470\, kg).(0.120\, m/s)^{2}}{18.5\, N/m}}$

Therefore:

${\color{Blue} x=\pm \, 0.0231\, m\approx 2.31\, cm}$