Recent content by DR1

Ok thank you for that, in that case my answer is that w = 119.99 rads. And in hertz I make the frequency 19.097? Thanks again for your assistance

and then dividing that by 2π gives me the answer in hertz?

-0.2838? then dividing by 0.05 gives -5.676 as my answer?

(73.74-90)/0.05 = -325.2 alternatively if I do the rads option (having now found a guide for my calculator!) i get (1.287002218-(π/2))/0.05 = 16.38 am i right in concluding that the first answer is in degrees and the second one is in rads so its the second one that I want?

Am i right in say that my calculator is working out the arcsine in degrees as the result I get is 73.74?

I am really struggling with this, my maths skills have never been particularly great as I am sure I am demonstrating! ω*0.05+90=arcsin(0.0192/0.02) You stated the above, to get w on its own I cannot work out whether I group up 0.05+90 and then put it in the denominator, or if I move both...

alternatively if i do ( Arcsin(0.0192/0.02))/90.05 = 0.82rads-1 this sounds better? could you confirm either way

ok so reaaranging for w i am getting (arcsin(0.0192/0.02)/0.05)-90=1384.796 this sounds to big to me were am i going wrong

δ=90 if i were to then put this with the figures gained once released 0.0192=0.02sin(w*0.05+90) how do i get w on its own? do i need to mutiply out the barckets

ok so i am still mot following I am afraid i am using the figures as displacement (x) = 0.020 time (t)=0 Amplitude (A)=0.020 initial Angle (O)= 0 are these the correct figures to use or am i best using the figures once released ie x=0.0192 t=0.05 A=0.02 O=0 i am trying to reaarange...

ok so A is amplitude which is the maximum displacement = 0.02m so putting all the info in gets 0.020=0.020sin(w0+angle) therefore 0.02/0.02=sinw which is 0.017 this does not look right to me i am obviously missing something please help

1. Homework Statement a mass attached to the lower end of a vertical spring causes the spring to extend by 25mm to its equilibrium position. The mass is then displaced a further 20mm and released. a trace of the vibration and time mesurements are taken. From these mesurements it can be seen...

so for the mass i reaarange I=mk^(2) so m=I/k^(2) =86.27/0.8^(2) m=134.8kg that sound ok

putting in all info into Ia(W1)a + Ib (W1)b=Ia(W2)a + Ib(W2)b 144a(15.71)b + Ib (8.37)b=144a(12.96)a +Ib (12.96)b 2262.24+Ib(8.37)b=1866.24+Ib(12.96)b 2262.24-1866.24 + Ib(8.37)b=Ib(12.96)b 396=Ib (12.96)-Ib(8.37) 396=Ib(4.49) Ib= 396/4.59 Ib= 86.27kg m^(2)

ok if i did it correctly i got an answer of Ib=86.27 but i not convinced my math is right