Physics Questions and Answers | Acceleration and Mass

QUESTION 5
Q5A) substitute D operator for dy
dx ,: (aD2+bD+C)x=0
Substituting m for D: am2+ bm+C=0
m= −b ± √ b2−4 ac
2a but r = -b/2a and ω= √4 ac−b2
2 a hence, jω = √b2−4 ac
2 a ,since b2 ≠ 4ac
hence, m= r±jω, ( complex roots). Therefore x(t) = ert(C1cos ωt+ C2sin ωt).
Q5B) (aD2+K)x=0
Substituting C for D: mC2+K=0; C= ± √−k
m , Let C= ±jω (since the C is complex)
Therefore, x(t)= e0(Acos ωt + Bsin ωt); Hence x(t) = Acos ωt + Bsin ωt
Q5C) x(t) = 0Acos ωt + Bsin ωt
Assume the initial conditions: y(0) =4, dx
dt = 9,
At x(0) =4, 4=Acos 0 + Bsin 0 i.e. A=4..........(i)
dx
dx = -ωAsinωt+ Bcos ωt; 9= 0 + Bcos 0; B=9......(ii)
Hence x(t)= 4cos ωt + 9 sin ωt
Let m=N7+1=3+1=4, and k= N8+1=9+1=10,
Hence, (4D2+10)X= 0;
Replacing D with C; 4C2+10=0; C=± √ −10
4 = ± j 1.58
X(t)= 4cos 1.58t +9sin 1.58t
Table 1: X(t)=4cos 1.58t + 9sin 1.58t
t in
seconds(s)
0 1.0 2.0 3.0 4.0 5.0 6.0
X in
metres
(m)
4 8.96 -4.17 -8.89 4.33 8.81 -4.49

0 1 2 3 4 5 6 7
-10
-8
-6
-4
-2
0
2
4
6
8
10
x
Figure1: plot for x(t)=4cos1.5t +9sin1.5t
Q5D) The product of acceleration and mass is the force producing the acceleration.
If all other factors are kept constant, mass is inversely proportional to time.
Q5 E) (mD2+qD+k)x=0
Replace D with C; mC2+qC+K=0
C= q ± √q 2−4 mk
2 m , where q2 -4mk< 0
Let r = −q
2m , and √q 2−4 mk
2 m = jω; so, C= r+jω
x(t)= ert(Acos ωt + Bsin ωt), A and B are constants.
Q5F) dx
dt = ertsin ωt(ωA+Br)+ = ertcos ωt(ωB+Ar)
9=Ar+ωB......(1)
4=e0(Acos0+Bsin 0); A=4
Since q=√4 mk-1, k=10 and m=4;
9=−6 √36+B √−1
8 ; B= -108; X(t)=4cos7.12t-108sin7.12t
Table2: X(t)=4cos7.12t-108sin7.12t
T(seconds) 0.5 1.0 1.5 2.0 2.5 3.0
X (metres) 40.48 -77.51 101.43 -107.84 195.65 104.15