Respuesta :
Answer:
[tex]K_2=\dfrac{K_1}{3}[/tex]
Explanation:
When only one spring is connected with mass
[tex]\omega _n=\sqrt{\dfrac{K_1}{m}}[/tex]
When another spring is added in series with spring one ,then the equivalent spring constant is K
Given that final natural frequency become half
[tex]\omega_f=\sqrt{\dfrac{K}{m}}[/tex]
[tex]\omega_f=\dfrac{1}{2}\omega _n[/tex]
[tex]K=\dfrac{1}{4}K_1[/tex] -----(1)
We know that equivalent spring constant for series connection is given as
[tex]K=\dfrac{K_1K_2}{K_1+K_2}[/tex] -----(2)
By using equation 1 and 2 we can say that
[tex]K_2=\dfrac{K_1}{3}[/tex]
Answer:
[tex]k_{2} = \frac{1}{3}k_{1}[/tex]
Solution:
Natural frequency of a spring mass system is given by:
[tex]\omega = \sqrt{\frac{k}{m}}[/tex] (1)
Now, with a system with [tex]k_{1}[/tex] and mass m, natural frequency is:
[tex]\omega_{n} = \sqrt{\frac{k_{1}}{m}}[/tex] (2) (given)
Also, when another spring [tex]k_{2}[/tex] is added in series with the first one the natural frequency of the system reduces to [tex]\frac{\omega_{n}}{2}[/tex], spring's equivalent stiffness is given by:
[tex]\frac{1}{k_{eq}}= \frac{1}{k_{1}} + \frac{1}{k_{2}}[/tex]
[tex]k_{eq} = \frac{k_{1}k_{2}}{k_{1} + k_{2}}[/tex]
Therefore,
[tex]\frac{\omega_{n}}{2} = \sqrt{\frac{k_{eq}}{m}}[/tex]
[tex]\omega_{n} = 2\sqrt{\frac{k_{eq}}{m}}[/tex] (3)
From eqn (2) and (3):
[tex]\sqrt{\frac{k_{1}}{m}} = 2\sqrt{\frac{k_{eq}}{m}}[/tex]
[tex]\sqrt{\frac{k_{1}}{m}} = 2\sqrt{\frac{k_{1}k_{2}}{m(k_{1} + k_{2})}}[/tex]
Squaring both sides of the above eqn, we get:
[tex]\frac{k_{1}}{m} = 4\frac{k_{1}k_{2}}{m(k_{1} + k_{2})}[/tex]
[tex]k_{1}^{2} = 4k_{1}k_{2} - k_{1}k_{2} = 0[/tex]
Solving the above equation in order to get the relation between [tex]k_{1}[/tex] and [tex]k_{2}[/tex]:
[tex]k_{1} = 3k_{2}[/tex]
Therefore, [tex]k_{2}[/tex] in terms of [tex]k_{1}[/tex]:
[tex]k_{2} = \frac{1}{3}k_{1}[/tex]
