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

a). C = b/2 and C = b/4

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

c). m = 63.4 kg (approx.)

Step-by-step 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}+ (k)/(m)x=0 $` or
`$ {\overset{..}x} + w_0^2 x = 0 $` , where
`$ w_0 = \sqrt{(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 = √(k/m) $`

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

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

a).
`$ m = (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{(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 = (b)/(2m_1) = (b)/(2m_2) $`

or
`$ \gamma = (b)/(2 * 1) = (b)/(2 * 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 = (k)/(m) \Rightarrow (k)/(w_0^2) = (k)/((2 \pi f_0)^2) = (k)/(4 \pi^2 f_0^2) $`

For two particle problem,

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

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

where, μ is the reduced mass.

This time period is same for both the particles.

c).
`$ m =(k)/(4 \pi^2 f_0^2)$`

`$ = (5 * 10^2)/(4 * 9.14^2 * 0.2) = 63.4\ kg $` ( approx.)