Please help with this problem ive got

A mass of 0.3 kg is suspended from a spring of stiffness 200 N m–1. If the mass is displaced by 10 mm from its equilibrium position and released, for the resulting vibration, calculate:

(i) the frequency of vibration

(ii) the maximum velocity of the mass during the vibration

(iii) the maximum acceleration of the mass during the vibration

(iv) the mass required to produce double the maximum velocity calculated in (ii) using the same spring and initial deflection.

Re: Please help with this problem ive got

You posted this in the algebra forum but I'm guessing that actually you are in a class that uses calculus, right? First set up the differential equation for a spring-mass system. From F=ma you have kx = ma, and since a = second derivative of x with respect to time, you have the differential equation:

$\displaystyle m \ddot x -kx = 0$

The general solution to this is $\displaystyle x = A \sin(\omega t + \alpha) $, where$\displaystyle A$ is amplitude, $\displaystyle \omega$ is rotatonal velocity in radians/second and $\displaystyle \alpha$ is the phase angle at t=0. From initial conditions you know A = 10mm and $\displaystyle \alpha$ is $\displaystyle \pi/4$ - do you see why that is? Now solve the initial differential equation for $\displaystyle \omega$ in terms of m and k. From here you can solve the first three questions. The final question will be apparent once you have the expression for $\displaystyle \omega$. Hope this helps.

Re: Please help with this problem ive got

I need to correct an error in my previous response. The spring force is F= -kx (I had +kx), and from F=ma this makes the basic equation ma+kx = 0. The correct differential equation is then:

$\displaystyle m \ddot x +kx = 0$

And just to be clear, the "double dot" notation means the 2nd derivative with respect to time:

$\displaystyle \ddot x = \frac {d^2x}{dt^2}$

Re: Please help with this problem ive got

sorry for late response busy with work and family

so far I have....

a) f = (1/2pi) x (square root k/m)

f = (1/2pi) x (square root 200/0.3)

f = 0.16 x 25.82

f = 4.1Hz

not sure from here on in sorry

Re: Please help with this problem ive got

OK, so far so good. You have $\displaystyle \omega = \sqrt{k/m} = 25.82 s^{-1}$, and so the equation of motion is

$\displaystyle x = A \sin(\omega t + \phi)$

You need to determine the amplitude of motion A and the initial phase angle $\displaystyle \tex\phi$, which you can get from the initial conditions of the mass's displacement at t=0.

The equation for velocity is the derivative of this:

$\displaystyle v = \frac {dx} {dt} = \frac {d(A \sin(\omega t + \phi))}{dt} = A \omega \cos (\omega t + \phi)$

From this, what's the max value for v?

Similarly the equation for acceleration is the derivative of velocity with respect to time.

For (iv) you'll have to determine a new value for $\displaystyle \omega$, and from that a value for m from $\displaystyle \omega = \sqrt {k/m}$. Hope this helps.