# Math Help - Simple Harmonic Motion

1. ## Simple Harmonic Motion

I can't do this question about SHM

A particle of mass 1.5kg is hanging in equilibrium attached to the free end of an elastic spring of natural length 1.2m which is hanging vertically. The particle receives an impulse of 6Ns parallel to the spring.
a) Show that the subsequent motion is simple harmonic. I have no idea where to being with this one.
b) Given that the period of motion is $\
\frac{{2\pi }}{5}
\$
calculate the modulus of the spring and the amplitude of the motion. This part of the question I used $\
v^2 = \omega ^2 (A^2 - x^2 )
\$
to find that the amplitude A=0.8m, but then I can't find the modulus of the spring $\
\lambda
\$
because I don't know the extension, so I'm stuck

Any help would be greatly appreciated.

2. Originally Posted by free_to_fly
I can't do this question about SHM

A particle of mass 1.5kg is hanging in equilibrium attached to the free end of an elastic spring of natural length 1.2m which is hanging vertically. The particle receives an impulse of 6Ns parallel to the spring.
a) Show that the subsequent motion is simple harmonic. I have no idea where to being with this one.
If the spring is light the impulse is the change in momentum of the mass, so
after the impulse the the velocity of the mass is 4 m/s up (or down the
question is ambiguous on this point). So you have an initial speed of 4 m/s,
and you proceed as usual with spring mass problems.

RonL

3. I've found that at equilibrium Tension T is given by $\
T = mg = \frac{{\lambda e}}{l}
\$
, where $\
\lambda
\$
=modulus of elasticity, e is the extension at equilibirum and l is the natural length.

So $\
\lambda e
\$
=17.64 (1.5*9.8)

Then using F=ma I got:
$\
9.8 - \frac{{\lambda (e + x)}}{{1.8}} = - \ddot x
\$
where x is further extension.
How do I use the velocity? I seem to have too many unknowns in this equation.

4. Originally Posted by free_to_fly
I've found that at equilibrium Tension T is given by $\
T = mg = \frac{{\lambda e}}{l}
\$
, where $\
\lambda
\$
=modulus of elasticity, e is the extension at equilibirum and l is the natural length.

So $\
\lambda e
\$
=17.64 (1.5*9.8)

Then using F=ma I got:
$\
9.8 - \frac{{\lambda (e + x)}}{{1.8}} = - \ddot x
\$
where x is further extension.
How do I use the velocity? I seem to have too many unknowns in this equation.
Solve the ODE with initial conditions $x(0)=0,\ \dot{x}(0)=4$.

By the way the ODE is:

$1.5 \times \ddot{x}=-\lambda x$

(force=mass times acceleration =mass times (second derivative of additional extension) = -(additional extension) times lambda

RonL

5. Oh no I'm confused now. Where did you get that ODE equation from? I thought the resultant force was weight-tension?

6. Originally Posted by free_to_fly
Oh no I'm confused now. Where did you get that ODE equation from? I thought the resultant force was weight-tension?
The equilibrium tension and extension should disappear if you go through the
problem in detail. But we normaly don't, we just work with the deviation from equilibrium.

RonL

7. I assume that your studying for your Mechanics 3 exam. you dont need to solve the ODE to show that the motion is harmonic your expected to be fimmilar with the standard result for S.H.M being $a = - \omega^2 x$ and use $\omega = \frac{2 \pi}{T}$ to get the time period

8. *nods* Yep, I'm doing M3 at the moment, how did you know that?
Anyway it's getting quiet confusing because I'm mixing it up with the formulae we use in physics for SHM, which are more simplified but they don't always work. But thanks for the help. Hopefully I can do more question now.

9. You have a british flag icon, you have posted nothing but FP2 and M3 questions. I"m also in year 13, looks like we are both in the same boat.

and the formula you use in physics are not differnet, they just use $k = \frac{ \lambda }{l}$

10. Great powers of deduction lol. FP2 and M3 are a major headache, and FP3's not that nice either.

I meant that in physics we only use $\
x = A\cos (2\pi ft)
\$
but in maths there's three: $
\
\begin{array}{l}
x = A\cos (\omega t) \\
x = A\sin (\omega t) \\
x = A\sin (\omega t + \alpha ) \\
\end{array}
\$

That initially confused me. It seems like all the calculations we do in physics are easier than they are in maths.

11. Originally Posted by free_to_fly
Great powers of deduction lol. FP2 and M3 are a major headache, and FP3's not that nice either.

I meant that in physics we only use $\
x = A\cos (2\pi ft)
\$
but in maths there's three: $
\
\begin{array}{l}
x = A\cos (\omega t) \\
x = A\sin (\omega t) \\
x = A\sin (\omega t + \alpha ) \\
\end{array}
\$

That initially confused me. It seems like all the calculations we do in physics are easier than they are in maths.
Dont worry we can make it through together

I guess you're doing edexcel PHY4. The use of calculus and trigonometric manipulation is not required for physics so the math will seem easier.

you applying for math at uni as well ?

12. Lol , hope we get through further maths in one piece. I've nearly finished FP2, thank god, but just starting FP3 bring it on lol.

I'm doing AQA syllabus, specification A for physics and I'm going to do physical natural sciences at uni. You doing maths? I did think about maths, but the thought of STEP changed my mind.

13. Originally Posted by free_to_fly
Lol , hope we get through further maths in one piece. I've nearly finished FP2, thank god, but just starting FP3 bring it on lol.

I'm doing AQA syllabus, specification A for physics and I'm going to do physical natural sciences at uni. You doing maths? I did think about maths, but the thought of STEP changed my mind.
So your a Cambridge applicant as well. I have a friend who is applying for the same course.
Talking about STEP I should go revise.