Initial displacement in Simple Harmonic Motion

[heartyface]

Homework Statement

A meterstick is clamped to a tabletop. The end of the meter stick is deflected downwards a small distance x and is released such the end of the meterstick moves up and down in simple harmonic motion. The meterstick is measured to oscillate up and down 10 times in 5.0 seconds. A very small mass m is ten placed on top of the end of the meterstick. The end of the meterstick is then deflected downwards a distance A and is released.

What is the minimum initial displacement A so that the small mass barely loses contact with the surface of the meterstick?

Homework Equations

T=2pi sqrt(L/g) ... because you can't use 2pi sqrt (k/m) because mass is not given?

The Attempt at a Solution

 

[tms]

You need to do more than just draw a picture. There are two parts of the problem. First, you use the information given to find out about the system without the added mass.

 

[heartyface]

so I did- I used T=0.5s=(2pi)sqrt(L/g)
and I got the 'acceleration' is 157.91...
If I use 0.5s=(2pi)sqrt(k/m) I cannot retrieve anything because m is too small...

 

[tms]

so I did- I used T=0.5s=(2pi)sqrt(L/g)

Where does L come from? Is the setup in the problem a pendulum?

and I got the 'acceleration' is 157.91...
If I use 0.5s=(2pi)sqrt(k/m) I cannot retrieve anything because m is too small...

It should be m/k, not k/m.

 

[heartyface]

aha, sorry. k/m --> m/k was a silly mistake of my fault.

Though this is not a pendulum problem at all, out of m/k and L/g I want to use an equation that does not involve m.

I think for in order to let the little m not slip off, the maximum acceleration of the meterstick cannot be greater than that of gravity.
To figure out the acceleration, I would love to use -kx=ma so a=-kx/m, but the lack of m value forbids me from doing so.
Ahh, I'm in a dilemma..

 

[tms]

Though this is not a pendulum problem at all, out of m/k and L/g I want to use an equation that does not involve m.

But you can't make stuff up; you have to use the correct equation. You also have to use the correct equation correctly; the added mass m is not the m that you are trying to plug into the equation, for instance.

Start with what you are given: that the end of the ruler executes SHM. Write down the general equation for SHM; that is, write down the equation that gives the position of the end of the ruler as a function of time. Then apply what you have found to the new conditions. The added mass is "very small", so it can be ignored.

I think for in order to let the little m not slip off, the maximum acceleration of the meterstick cannot be greater than that of gravity.

Right.

 

[heartyface]

@tms, you have been helping me all throughout and instead of merely telling me the answer you gradually led me here.
I thank you so much, for I know I truly learned.
Because kx=mg, I can say m=kx/g
thus for 2pi sqrt(m/k) i can plug in 2pi sqrt(x/g)
thus 0.5=2pi (sqrt x/9.8) so x=0.062 or 6.2cm.
Thank you.

 

[tms]

You're welcome.