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Question

A vehicle suspension system consists of a spring and a damper. The stiffness of the spring is 3.6 kN/m and the damping constant of the damper is 400Ns/m if the mass is 50Kg find damping factor and damped natural frequency respectively are

The correct answer is

0.471 and 1.19 Hz

Understanding Vehicle Suspension Systems

A vehicle suspension system is a critical component designed to absorb shocks and vibrations, providing a smooth ride and maintaining tire contact with the road surface. It typically consists of a spring and a damper (also known as a shock absorber). The spring stores energy and absorbs shocks, while the damper dissipates energy, controlling oscillations.

Key Parameters of the Suspension System

In this problem, we are given the following parameters for the vehicle suspension system:

Parameter Symbol Value
Mass \(m\) \(50 \, \text{Kg}\)
Spring Stiffness \(k\) \(3.6 \, \text{kN/m} = 3600 \, \text{N/m}\)
Damping Constant \(c\) \(400 \, \text{Ns/m}\)

Calculating the Damping Factor

The damping factor, often denoted by \(\zeta\) (zeta), is a dimensionless measure describing how oscillations in a system decay after a disturbance. It is the ratio of the actual damping constant (\(c\)) to the critical damping constant (\(c_c\)).

First, we need to calculate the natural frequency (\(\omega_n\)) of the undamped system:

\[\omega_n = \sqrt{\frac{k}{m}}\]

Substituting the given values:

\[\omega_n = \sqrt{\frac{3600 \, \text{N/m}}{50 \, \text{Kg}}} = \sqrt{72} \, \text{rad/s}\]

\[\omega_n \approx 8.485 \, \text{rad/s}\]

Next, we calculate the critical damping constant (\(c_c\)), which is the amount of damping required for the system to return to equilibrium as quickly as possible without oscillating:

\[c_c = 2 \sqrt{km} = 2m\omega_n\]

Using \(2m\omega_n\):

\[c_c = 2 \times 50 \, \text{Kg} \times 8.485 \, \text{rad/s} = 100 \times 8.485 \, \text{Ns/m} \approx 848.5 \, \text{Ns/m}\]

Alternatively, using \(2 \sqrt{km}\):

\[c_c = 2 \sqrt{3600 \, \text{N/m} \times 50 \, \text{Kg}} = 2 \sqrt{180000} \, \text{Ns/m} \approx 2 \times 424.26 \, \text{Ns/m} \approx 848.52 \, \text{Ns/m}\]

Now we can calculate the damping factor (\(\zeta\)):

\[\zeta = \frac{c}{c_c}\]

Using \(c = 400 \, \text{Ns/m}\) and \(c_c \approx 848.52 \, \text{Ns/m}\):

\[\zeta = \frac{400}{848.52} \approx 0.4714\]

So, the damping factor is approximately 0.471.

Calculating the Damped Natural Frequency

The damped natural frequency (\(\omega_d\) or \(f_d\)) is the frequency at which a damped system oscillates when disturbed from its equilibrium position. For an underdamped system (\(\zeta < 1\)), the damped natural frequency (\(\omega_d\)) in rad/s is related to the natural frequency (\(\omega_n\)) and the damping factor (\(\zeta\)) by the formula:

\[\omega_d = \omega_n \sqrt{1 - \zeta^2}\]

We found \(\omega_n \approx 8.485 \, \text{rad/s}\) and \(\zeta \approx 0.4714\). Substituting these values:

\[\omega_d = 8.485 \sqrt{1 - (0.4714)^2}\]

\[\omega_d = 8.485 \sqrt{1 - 0.2222}\]

\[\omega_d = 8.485 \sqrt{0.7778}\]

\[\omega_d = 8.485 \times 0.8819\]

\[\omega_d \approx 7.484 \, \text{rad/s}\]

The question asks for the frequency in Hertz (Hz). The relationship between angular frequency (\(\omega_d\) in rad/s) and frequency (\(f_d\) in Hz) is \(f_d = \frac{\omega_d}{2\pi}\).

\[f_d = \frac{7.484 \, \text{rad/s}}{2\pi}\]

\[f_d = \frac{7.484}{6.283} \, \text{Hz}\]

\[f_d \approx 1.191 \, \text{Hz}\]

So, the damped natural frequency is approximately 1.19 Hz.

Summary of Results

  • Damping Factor (\(\zeta\)): Approximately 0.471
  • Damped Natural Frequency (\(f_d\)): Approximately 1.19 Hz

These values represent how the suspension system will behave when subjected to disturbances. A damping factor between 0 and 1 indicates an underdamped system, meaning it will oscillate with decreasing amplitude over time at the damped natural frequency.

Based on our calculations, the damping factor is approximately 0.471 and the damped natural frequency is approximately 1.19 Hz. These values match the first option provided.

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Important Questions from Damped Free Vibration

  1. When there is reduction in amplitude over every cycle of vibration, then the body is said to have

  2. Which condition is suitable for indicating instruments in order to get the best results?

  3. A single degree of freedom system, having mass of 1 kg and stiffness of 10 kN/m is at rest. It is subjected to an impulsive force of magnitude 5 kN for 10-4 seconds. The amplitude (in mm) of the resulting free vibration is

  4. Which of the following statements are TRUE for damped vibrations?

    P. For a system having critical damping, the value of the damping ratio is unity and the system does not undergo a vibratory motion.

    Q. Logarithmic decrement method is used to determine the amount of damping in a physical system.

    R. In case of damping due to dry friction between moving surfaces resisting force of constant magnitude acts opposite to the relative motion.

    S. For the case of viscous damping, drag force is directly proportional to the square of relative velocity.

  5. A suspended mass of 10 kg completes 40 oscillations in 20 seconds in a single-degree damped vibrating system. The stiffness of the spring is approximately _________.

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