- #1

wezzo62

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It's a mass spring system problem and I am just not sure if I am doing it right or not. If you could have a look at the problem attached and my solutions below and just let me know if I am doing it right or not

1)a) k

b) k

c) mx" + kx = 0 => 6x" + 2355x = 0 (eq. of motion)

x = sin(ω

x' =Aωcos(ω

x" = -ω

then substituting (1) and (3) back into eq. of motion and dividing by

Asin(ω

gives: -mω

d) same process as for c) and got: ω

Im now a little stuck/confused on e) and f) and its making me think what I've done so far is wrong! this is really annoying me so please could someone tell me if any of this is right.

2)a) log decrement, δ = ln(10/5) =

b) damping factor, ζ = δ/2pi =

c) natural frequency, ω = √(76/4) = √19

=> damped natural frequency, ω

d) c

=> damping constant, c = ζ c

I think most of Q2 should be right . . ? any help much appreciated, thanks

1)a) k

_{eq}= k_{1}+ k_{2}=__2355N/m__b) k

_{eq}= 1 / ((1/k_{1}) + (1/k_{2})) =__451.0301N/m__c) mx" + kx = 0 => 6x" + 2355x = 0 (eq. of motion)

x = sin(ω

_{n}t + ∅) ___________________________________(1)x' =Aωcos(ω

_{n}t + ∅) _________________________________(2)x" = -ω

_{n}^{2}Asin(ω_{n}t + ∅) __________(3)then substituting (1) and (3) back into eq. of motion and dividing by

Asin(ω

_{n}t + ∅)gives: -mω

_{n}^{2}+ k = 0 => ω_{n}= √(k/m) =__19.81 rad/s__d) same process as for c) and got: ω

_{n}= √(451.0301/6) =__8.67rad/s__Im now a little stuck/confused on e) and f) and its making me think what I've done so far is wrong! this is really annoying me so please could someone tell me if any of this is right.

2)a) log decrement, δ = ln(10/5) =

__0.6931__b) damping factor, ζ = δ/2pi =

__1.0888__c) natural frequency, ω = √(76/4) = √19

=> damped natural frequency, ω

_{d}= ω√(1-ζ^{2}) =__4.3323__d) c

_{cr}= 2√(km) = 11.1335 (bearing in mind its 4N not 4kg and therefore m=4/9.81=0.41kg)=> damping constant, c = ζ c

_{cr}=__1.2282__I think most of Q2 should be right . . ? any help much appreciated, thanks

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