# The required mass to produce double the maximum velocity.

## Homework Statement

￼3. 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:
(a) (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.

(b) Plot a graph of acceleration against displacement (x) (for values of x
from x = –10 mm to x = +10 mm)

## Homework Equations

(a)
(i).. i used equation.. frequency=1/2pi*sqrt 200/0.3 = 4.11hz

(ii) to get angular frequency(w)... w=sqrt k/m ---> w= sqrt 200/0.3 = 25.82 rads

i used equation.. v=A*wcos(w*t+starting angle)---> 0.01*25.82cos(25.82*0+0)
v= 0.01*25.82cos ----> v= 0.2582 ms or 0.26 ms

(iii) equation used... a=A*w^2sin(w*t+starting angle) -->
a= 0.01*(25.82)^2sin(25.82*0+0)
a=0.01*(25.82)^2sin ----> a= 6.667 ms

(iv) i am unsure on how to calculate this can anyone offer any help with an equation to work it out so i can obtain the answer myself.

(b) i have 6.667 at -10mm and -6.667 at +10mm... i know they are correct but how can i calculate points in between like +/-5mm to get it more accurate..

## The Attempt at a Solution

any help would be great but am not looking for the answers just an equation to help me

## Answers and Replies

for question (iv) i have done v*2= 0.2582*2=0.5164

if i do mass required for double velocity= m*v--> 0.3*0.2582=0.077 kg m
so if i do angular frequency (w)= sqrt k/m ----> sqrt 200/0.077 = 50.96

v= 0.01*50.96= 0.5096 which does not equal 2*v which is 0.5164

but if i do m*v and use 0.3*0.25= 0.075 kg m

so (w) = sqrt k/m ---> sqrt 200/0.075= 51.64

so v= 0.01*51.64= 0.5164 which equals the v*2 of 0.5164

which one am i doing correct or are they both wrong because i cant think of any way of doing it.

rude man
Homework Helper
Gold Member
"Starting angle"? Why do you assume the mass is displaced angularly? I read it as being displaced vertically. Can't be angular because the length of the spring is not given ....

"Starting angle"? Why do you assume the mass is displaced angularly? I read it as being displaced vertically. Can't be angular because the length of the spring is not given ....

It's how it words the symbol in my lesson books.. So I get v=A*wcos (wt+0)
(wt+0) is called phase (rad) in my book