Dark mode: OFF

# Damped free Vibration - Numerical 1

Comments

#### A mass of 85 kg is supported on a spring which deflects 18 mm under the weight of the mass. The vibrations of the mass are constrained to be linear and vertical. A dashpot is provided which reduces the amplitude to one-quarter of its initial value in two complete oscillations. Calculate magnitude of the damping force at unit velocity and periodic time of damped vibration.

Solution: Fig-1

Given data:

mass $\ m = 85$ kg
Static deflection, $\ x = 18\ \text{mm} = 0.018\ \text{m}$

Let,
$x_0 =$ initial amplitude
$x_2=$ final amplitude after two complete cycle $= {1\over4}$

Taking ratio,

${x_0 \over x_2}={x_0 \over {1 \over 4} x_0}=4$

#### Damping force:

$\text F = c.\mathring x= c$

Now,

\begin{aligned} \xi= {c \over c_c} \quad \text {OR} \quad &\xi= {c \over 2m\omega_0} \quad (\text {as} \quad c_c=2m\omega_n)\\ \therefore \ &c=(2m\omega_n)\xi \ \dots\dots\dots\dots \ (1) \end{aligned}

#### 1. The Logarithmic decrement:

\begin{aligned} \delta &= {1 \over n} log_e\bigg ({x_0 \over x_n}\bigg)\\ &= {1 \over 2} log_e\bigg ({x_0 \over x_2}\bigg)\\ &= {1 \over 2} log_e(4) \end{aligned}

#### 2. Damping factor:

\begin{aligned} \xi &= {\delta \over \sqrt{4 \pi^2 + {\delta}^2} } \\ &={\delta \over \sqrt{4 \pi^2 + {\delta}^2} }\\ &= 0.1091 \end{aligned}  #### 3. Frequency of undamped free vibration:

The natural circular frequency of vibration is,

\begin{aligned} \omega_n &= {\sqrt{k \over \ m}} \\ &= {\sqrt{m \ g \over \ x \ m}}\\ &= {\sqrt{g \over x}} \\ &= {\sqrt{9.81 \over 0.018}}\\ &= 23.34 \ \text {rad/s} \end{aligned}

From equation (1); we get,
Damping force at unit velocity;

\begin{aligned} \text F = c &= (2m\omega_n)\xi \\ &= (2 \ \text x \ 85\ \text x \ 23.34)\ \text x \ (0.1091) \\ &= 91.67 \ \text N \end{aligned}

The time period of damped vibration is,

\begin{aligned} t_p &= {2\pi \over \omega_d} \\ &= {2\pi \over \omega_n \ \sqrt {1 -\xi^2}} \\ &= {2\pi \over (23.34) \ \sqrt {1 -(0.1091)^2}} \\ &=0.2708 \ \text {sec} \end{aligned}

### Help us build Education Lessons

If our notes and videos are helpful to you, kindly support us by making a donation from our support page so we can continue making more content for students like you.

Go to support page

### Comments:

All comments that you add will await moderation. We'll publish all comments that are topic related, and adhere to our Code of Conduct.

Want to tell us something privately? Contact Us

No comments available.