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Old Jul 23rd 2019, 08:00 PM   #1
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Spring related problem, force time mass and displacement

Hello,

Kindly solve this problem for me, or at least give me the procedure of equation to use. I don't know how to begin because I have not found any relationship that relate spring constant to time. If you find the solution, give me numerical result.

We have (see the picture):

1. A Compression spring fixed on one side to a reference. The spring free length is L.
2. An object of mass (m) attached to the spring.
3. A wall on which the mass (m) will be hitting.

Initially, the spring is at position (1) and the mass (m) have no velocity. When Force holding (m) disappear, The spring will push the mass (m) toward the wall with hatching lines. The distance traveled by the mass (m) is $\displaystyle x_2-x_1$ and this distance took a duration of time equal to t.

- Spring free length: L
- $\displaystyle F_1$: Force of spring when compressed by displacement $\displaystyle x_1$
- $\displaystyle F_2$: Force of spring when compressed by displacement $\displaystyle x_2$

Data available:
- Mass : m = 125g = 0.125 Kg
- Duration : t = 0.6 s
- Force : $\displaystyle F_1$ = 15 Newtons
- $\displaystyle x_2-x_1$ = 3 cm = 0.03 meter

Questions:
1- What is the amplitude of the Force $\displaystyle F_2$ ?
2- What is the spring constant $\displaystyle K$ ?


Thank You.
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Old Jul 24th 2019, 04:43 AM   #2
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An ideal spring has some "spring constant", K, such that the force necessary to extend the spring distance x beyond its natural length (or compress it to distance x from its natural length) is F= Kx.

If I am reading this correctly, initially a force, $\displaystyle F_2$, is compressing the spring a distance $\displaystyle x_2$ inside its natural length (this is ambiguous because you say the natural length is "L" but in your picture you have "l"). If this is correct then $\displaystyle F_2= Kx_2$.

When that force is "released" the spring will react with $\displaystyle ma= Kx_2$ where a is the acceleration of the mass. $\displaystyle ma= m\frac{dv}{dt}= Kx_2$ so $\displaystyle mv(t)= m\frac{dx}{dt}= Kx_2t$ where v(t) is the velocity of the mass at time t after being released. The distance the mass will travel in time t is $\displaystyle x(t)= \frac{Kx_2}{2m}t^2$.

The mass will have traveled distance $\displaystyle x_2- x_1$ when $\displaystyle \frac{Kx_2}{2m}t^2= x_2- x_1$. $\displaystyle t^2= \frac{2m(X_2- X_1)}{Kx_2}$ so $\displaystyle t= \sqrt{\frac{2m(X_2- X_1)}{Kx_2}}$.
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Last edited by HallsofIvy; Jul 28th 2019 at 05:06 AM.
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