1. ## Frequency and Tension

A cable is used to suspend a 800kg safe. The cable is elastic and stretches 20mm when subjected to a tension of 4kN.

a) Find the stiffness of the cable, and it's extension when the safe is hanging in equlibrium.

The safe is being lowered at 6m/s when the motor controlling the cable suddenly jams.

b) Determine the frequency with which the safe vibrates on the end of the cable.

c) Determine also the maximum tension in the cable during the oscillations.
Right sorry for the long question, here's what I've done so far.

a) In the initial state, $\displaystyle T = ka$ where $\displaystyle a$ is the extension and $\displaystyle k$ is the stiffness constant.

Putting our initial conditions into the above equation, we get $\displaystyle 4000 = 0.02k$, so the stiffness $\displaystyle k = 200,000$

To find the extension when hanging in equilbrium, we use the fact that $\displaystyle T = ka = mg$.

When the safe is hanging on the cable, this becomes $\displaystyle 200000a = 800g$, so the extension $\displaystyle a = 0.0392m \approx 40mm$

b) Using Newton's Second Law, $\displaystyle ma = F$, resolvign vertically downwards, this gives us $\displaystyle m\ddot{x} = mg - T$

$\displaystyle m\ddot{x} = mg - k(x+a)$ where $\displaystyle x = x(t)$ is the extra extension due occurring with the oscillation motion.

$\displaystyle m\ddot{x} = mg - kx -ka$, now we know from above that $\displaystyle mg = ka$, so:

$\displaystyle m\ddot{x} + kx = 0$, $\displaystyle \ddot{x} + \frac{k}{m}x = 0$

Now if we let $\displaystyle \frac{k}{m} = \omega^2$, our DE becomes $\displaystyle \ddot{x} + \omega^2x = 0$, solving this gives us.

$\displaystyle x(t) = A\cos{\omega t} + B\sin{\omega t}$

and

$\displaystyle \dot{x}(t) = -\omega A\sin{\omega t} + \omega B\cos{\omega t}$

We know that $\displaystyle \dot{x}(0) = 6$, putting this into our second equation we get $\displaystyle 6 = \omega B$, $\displaystyle B = \frac{6}{\omega}$.

Little confused about the initial conditions for $\displaystyle x(0)$ though, is the displacement zero, or is it the $\displaystyle a$ that I calculated in the first part of the question?

Also how does do I calculate the frequency from this equation?

c) Not sure how to start this one either, guessing it involves inequalities but apart from that not sure.

Thanks in advance for the help

2. maybe im misunderstanding, but you got

$\displaystyle \omega$ is the frequency. its done
notice that the frequency never depends on the initial speed

for c) =0
youll also need the initial speed condition

you can also do it by energy conservation

3. Originally Posted by Haytham
$\displaystyle \omega$ is the frequency. its done
notice that the frequency never depends on the initial speed
Of course!! Sorry brain just decided not to work then.

Originally Posted by Haytham
for c) =0
youll also need the initial speed condition

you can also do it by energy conservation
From the initial speed, you know $\displaystyle B = \frac{6}{\omega}$, so

$\displaystyle x(t) = A\cos{\omega t} + \frac{6}{\omega}\sin{\omega t}$

$\displaystyle x(0) = 0$, $\displaystyle 0 = A$, so the equation for $\displaystyle x(t)$ is:

$\displaystyle x(t) = \frac{6}{\omega}\sin{\omega t}$?

How would I go about finding the tension from here?

4. T=-k (x+a)

so the tension is maximum when x is maximum

find maximum of x

as
$\displaystyle x(t) = \frac{6}{\omega}\sin{\omega t}$

maximum of x is
$\displaystyle \frac{6}{\omega}$

then replace in
T=-k (x+a)

5. Originally Posted by Haytham
T=-k (x+a)

so the tension is maximum when x is maximum

find maximum of x

as
$\displaystyle x(t) = \frac{6}{\omega}\sin{\omega t}$

maximum of x is
$\displaystyle \frac{6}{\omega}$

then replace in
T=-k (x+a)
Thank you!