Question

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's equilibrium location), velocity, and acceleration of the block are x = 0.134 m, v = -12.1 m/s, and a = -107 m/s2. Calculate (a) the frequency of oscillation, (b) the mass of the block, and (c) the amplitude of the motion.

Answer 1

a) 4.49Hz

b) 0.536kg

c) 2.57s

Step-by-step explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

`x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)`

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

`(x)/(a)=(Acos(\omega t))/(-\omega^2 Acos(\omega t))\\\\\omega=\sqrt{-(a)/(x)}=\sqrt{-(-107m/s^2)/(0.134m)}=28.25(rad)/(s)`

with this value you can compute the frequency:

a)

`f=(\omega)/(2\pi)=(28.25rad/s)/(2\pi)=4.49Hz`

b)

the mass of the block is given by the formula:

`f=(1)/(2\pi)\sqrt{(k)/(m)}\\\\m=(k)/(4\pi^2f^2)=(427N/m)/((4\pi^2)(4.49Hz)^2)=0.536kg`

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

`(v)/(x)=-\omega tan(\omega t)\\\\t=(1)/(\omega)arctan(-(v)/(x\omega ))=(1)/(28.25rad/s)arctan(-(-12.1m/s)/((0.134m)(28.25rad/s)))=2.57s`

Finally, the amplitude is:

`x=Acos(\omega t)\\\\A=(0.134m)/(cos(28.25rad/s*2.57s ))=0.45m`