Question

5. [1pt]
A mass is vibrating at the end of a spring with a spring constant 2.09 N/m. The figure shows a graph of its position x (in centimetres) as a function of time t (in seconds). At what time between t=0 s and the first maximum after t=0 s is the mass not moving?

ON

Correct, computer gets: 5.00e-01 s
6. [1pt]
What is the magnitude of the acceleration of the object at the second maximum in the x-t curve after t = 0 s?

Correct, computer gets: 1.34e+02 cm/s^2

7. [1pt]
What is the mass of the object?

8. [1pt]
How much energy did the system originally contain?

9. [1pt]
How much energy did the system lose between t = 0 s and the third maximum after t = 0 s? Think about where this energy has gone.

Answer 1

The general equation of damped SHM will be like (the mass m, and the force constant k, and the resistance force coefficient b)

$x\left ( t \right )=Ae^{-t/\tau }\cos\left ( wt-\delta \right )$

Where

$\tau =\frac{2m}{b},\ and\ \omega =\sqrt{\frac{k}{m}-\frac{1}{\tau ^{2}}}$

From the graph, the time period T = 1 s, hence the angular frequency is

$\omega =\sqrt{\frac{k}{m}-\frac{1}{\tau ^{2}}}=\frac{2\pi }{T}=2\pi$

7. Initial amplitude, A = 6 cm, after two time period (2T = 2 s), the amplitude is 3.4 cm. Hence from the above equation

$\frac{6}{3.4}=e^{2/\tau },\ or\ \tau =3.52\ s$

Therefore, from the equation for angular frequency

$\frac{k}{m}-\frac{1}{\tau ^{2}}=4\pi^{2}$

$\therefore m=\frac{k}{4\pi^{2}-\frac{1}{\tau ^{2}}}=\frac{2.09}{4\pi^{2}-\frac{1}{3.52 ^{2}}}=0.053 \ Kg$

Mass of the object is 0.053 Kg

8. The initial energy will be the potential energy as kinetic energy is zero (amplitude is 6 cm or 0.06 m)

$E=\frac{1}{2}m\omega ^{2}A^{2}=\frac{1}{2}\times 0.053\times 4\pi ^{2}\times 0.06^{2}=3.77\ mJ$

9. The amplitude is 2.6 cm or 0.026 m, the energy will be

$E=\frac{1}{2}m\omega ^{2}A^{2}=\frac{1}{2}\times 0.053\times 4\pi ^{2}\times 0.026^{2}=0.71\ mJ$

The energy lose by the system will be 3.77 - 0.71 or 3.06 mJ.