Question

Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object

Answer 1

Answer:

a). C = b/2   and C = b/4

b).  $\therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$

c). m = 63.4 kg (approx.)

Explanation:

Ex. 2.4.4

The total force acting on mass m is $F = F_{spring }= -kx$ , where x is the displacement from the equilibrium position.

The equation of motion is $m {\overset{..}x} + kx = 0$

or ${\overset{..}x}+ \frac{k}{m}x=0$     or   ${\overset{..}x} + w_0^2 x = 0$ ,    where $w_0 = \sqrt{\frac{k}{m}}$  

The solution is $x = A \cos (w_0t + \phi)$ ,  where A and Ф are constants.

A is amplitude of  motion

$w_0$ is the angular frequency of motion

Ф is the phase angle.

Now, $w_0 = 2 \pi f_0 = \sqrt{k/m}$

or $m = \frac{k}{4\pi f_0^2}$

Given $f_0 = 0.8 Hz , k = 4 N/m$

a).  $m = \frac{4}{4(3.14)^2(0.8)^2} = 0.158\ kg$

b). $w_0^2 = k/m$  

or  $m = k/ w_0^2 = k / (2\pi f_0)^2 = k / 4 \pi^2 f_0^2$

Ex. 2.4.5

a). Total force acting on the mass m is $F = F_{spring}+f$

                                                                    $= -kx-bv$

    The equation of motion is $m {\overset{..}x}= -kx-b{\overset{.}x}$

    or $w_0 = \sqrt{\frac{k}{m}}$ ,  angular frequency of the undamped oscillation.

   γ = b/2m  is called the damping coefficient (γ=C)

   $k = m w_0^2 = 4 \pi^2 m f_0^2$

   for 1 kg weight (= 9.8 N), $f_0$ = 1.1 Hz

    k = 4 x (3.14)² x (9.8) x 1.1²  = 4.6 x 10² N/m

  For 2 kg weight (= 19.6 N), $f_0$ = 0.8 Hz

    k = 4 x 9.8596 x 2 x 9.8 x 0.8²  = 5 x $10^7$ N/m

   $\gamma = \frac{b}{2m_1} = \frac{b}{2m_2}$

or $\gamma = \frac{b}{2 \times 1} = \frac{b}{2 \times 2}$

γ = b/2 (for 1 kg)  and   γ = b/4 (for 2 kg)

C = b/2   and C = b/4

b). $w_0^2 = \frac{k}{m} \Rightarrow \frac{k}{w_0^2} = \frac{k}{(2 \pi f_0)^2} = \frac{k}{4 \pi^2 f_0^2}$

 For two particle problem,

   $w'_0^2 = \sqrt{\frac{k(m_1+m_2)}{m_1 +m_2}}$

$\therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$

where, μ is the reduced mass.

This time period is same for both the particles.

c). $m =\frac{k}{4 \pi^2 f_0^2}$

        $= \frac{5 \times 10^2}{4 \times 9.14^2 \times 0.2} = 63.4\ kg$  ( approx.)