Question

Question: In Figure, the block has a mass of 1.50kgand the spring constant is800 N/m. The damping force is given by -b(dx/dt), where b = 230 g/s. The block is pulled down 12.0 cmand released.

1. Calculate the time required for the amplitude of the resulting oscillations to fall to one-third of its initial value.
2. How many oscillations are made by the block in this time?

Short answer

Answer

1. The time for the oscillator amplitude to become one-third of initial value is t = 14.3 s
2. The number of oscillations, n = 5.27

Step-by-step solution

Given

1. The mass of the block is, m = 1.5 kg
2. The damping constant of the oscillator is, b = 230 g/s = 0.23 kg/s
3. Spring constant, k = 800 N/m

The displacement of block, x = 12 cm = 0.12 m

Understanding the concept

Use the equation for the damping factor. By rearranging it for the time, find the required time. Then using the equation for the period, calculate the period of oscillation. The ratio of time required for the amplitude to fall to one-third of its initial value and the period will give us the number of oscillations.

The amplitude of oscillations after n cycles is given as-

${\mathbf{x}}_{\mathbf{n}}\mathbf{=}{\mathbf{x}}_{\mathbf{m}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{bt}}{\mathbf{2}\mathbf{m}}}$

The time period of oscillation in the case of spring is given as-

$\mathbf{T}\mathbf{=}\mathbf{2}\mathbf{\pi }\sqrt{\frac{\mathbf{m}}{\mathbf{k}}}$

Here, m is the mass of the pendulum, k is the force constant of the spring, b is damping constant and t is the time taken.

(a) Calculate the time required for the amplitude of the resulting oscillations to fall to one-third of its initial value

We calculate the time for which the damping factor is 1/3.

So,

$e^{-\frac{\text{bt}}{2\text{m}}}=\frac{1}{3}$

By rearranging this equation for time t, we get

$\text{t}=-\frac{2\text{m}}{\text{b}}\text{ln}\left(\frac{1}{3}\right)$

So, using the given values, we get

t = 14.3 s

(b) Calculate the number of oscillations made by the block in this time

Now, we have the equation for period of damped oscillation as

$\text{T}=\frac{2\pi }{\omega \text{'}}$

Where,is the angular frequency of damped oscillation and is given by

$$\begin{gathered} \omega \text{'}=\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}} \\ \sqrt{\frac{8.00N/m}{1.5\text{kg}}-\frac{{\left(0.23kg/s\right)}^2}{4\times {\left(1.5\text{kg}\right)}^2}} \\ =2.31rad/s \end{gathered}$$

Hence, the period of the damped oscillation becomes

$$\begin{gathered} \text{T}=\frac{2\pi }{\omega \text{'}} \\ =\frac{2\pi }{2.31rad/s} \\ =2.72\text{s} \end{gathered}$$

So, the number of oscillations are-

$$\begin{gathered} n=\frac{t}{T} \\ =\frac{14.3\text{s}}{2.72\text{s}} \end{gathered}$$

n = 5.27

The number of oscillations mad by the block is 5.27.