
A mass of \[5\,{\text{kg}}\] is suspended from a spring of stiffness \[46\,{\text{kN/m}}\] . A dashpot is fitted between the mass and the support with a damping ratio of \[0.3\] . Calculate the damped frequency.
A. \[14.56\,{\text{Hz}}\]
B. \[14.28\,{\text{Hz}}\]
C. \[14.42\,{\text{Hz}}\]
D. \[14.14\,{\text{Hz}}\]
Answer
586.2k+ views
Hint: First of all, we will find out the natural frequency of the system by substituting the required values. We will use this value and find out the damped frequency by using the formula of damped frequency which includes a physical quantity called damping factor.
Complete step by step answer:
In the given question, we are supplied the following data:
Mass of the object suspended is \[5\,{\text{kg}}\] .
Spring constant is given as \[46\,{\text{kN/m}}\] which can be written as \[46{\text{000}}\,{\text{N/m}}\] .
The damping ratio/factor is \[0.3\] .
We are asked to find the damped frequency. For this we need to first find the natural undamped frequency of the system.
The natural undamped frequency of the system is given by the formula given below:
\[{f_n} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{M}} \] …… (1)
Where,
\[{f_n}\] indicates the natural undamped frequency.
\[k\] indicates the spring constant.
\[M\] indicates the mass of the object.
Now we substitute the required values in the equation (1) and we get,
$ {f_n} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{M}} \\ $
$ {f_n} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{46000}}{5}} \\ $
$ {f_n} = \dfrac{1}{{2\pi }}\sqrt {9200} \\ $
$ {f_n} = 15.27\,{\text{Hz}} \\ $
The natural undamped frequency is found out to be \[15.27\,{\text{Hz}}\] .
Again, the damped frequency is given by the formula given below:
\[f = {f_n}\sqrt {1 - {\delta ^2}} \] …… (2)
Where,
\[f\] indicates the damped frequency.
\[{f_n}\] indicates the undamped frequency.
\[\delta \] indicates the damping ratio.
Now, we substitute the required values in the equation (2), we get:
$ f = 15.27\sqrt {1 - {{\left( {0.3} \right)}^2}} \\ $
$ f = 15.27\sqrt {1 - 0.09} \\ $
$ f = 15.27\sqrt {0.91} \\ $
$ f = 14.56\,{\text{Hz}} \\ $
Hence, the damped frequency is found out to be \[14.56\,{\text{Hz}}\] .
So, the correct answer is “Option A”.
Note:
It is important to note that the highest natural frequency of an object is always lowered by damping. If the amplitude reduces with time, then the oscillation is called damped oscillation. In case of undamped oscillation, the amplitude does not change with time. Remember that in case of damped oscillation, the frequency will decrease as compared to undamped one.
Complete step by step answer:
In the given question, we are supplied the following data:
Mass of the object suspended is \[5\,{\text{kg}}\] .
Spring constant is given as \[46\,{\text{kN/m}}\] which can be written as \[46{\text{000}}\,{\text{N/m}}\] .
The damping ratio/factor is \[0.3\] .
We are asked to find the damped frequency. For this we need to first find the natural undamped frequency of the system.
The natural undamped frequency of the system is given by the formula given below:
\[{f_n} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{M}} \] …… (1)
Where,
\[{f_n}\] indicates the natural undamped frequency.
\[k\] indicates the spring constant.
\[M\] indicates the mass of the object.
Now we substitute the required values in the equation (1) and we get,
$ {f_n} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{M}} \\ $
$ {f_n} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{46000}}{5}} \\ $
$ {f_n} = \dfrac{1}{{2\pi }}\sqrt {9200} \\ $
$ {f_n} = 15.27\,{\text{Hz}} \\ $
The natural undamped frequency is found out to be \[15.27\,{\text{Hz}}\] .
Again, the damped frequency is given by the formula given below:
\[f = {f_n}\sqrt {1 - {\delta ^2}} \] …… (2)
Where,
\[f\] indicates the damped frequency.
\[{f_n}\] indicates the undamped frequency.
\[\delta \] indicates the damping ratio.
Now, we substitute the required values in the equation (2), we get:
$ f = 15.27\sqrt {1 - {{\left( {0.3} \right)}^2}} \\ $
$ f = 15.27\sqrt {1 - 0.09} \\ $
$ f = 15.27\sqrt {0.91} \\ $
$ f = 14.56\,{\text{Hz}} \\ $
Hence, the damped frequency is found out to be \[14.56\,{\text{Hz}}\] .
So, the correct answer is “Option A”.
Note:
It is important to note that the highest natural frequency of an object is always lowered by damping. If the amplitude reduces with time, then the oscillation is called damped oscillation. In case of undamped oscillation, the amplitude does not change with time. Remember that in case of damped oscillation, the frequency will decrease as compared to undamped one.
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