SOLUTIONS. 19.If x=,f. Solution. But x=. Replacing in.

SOLUTIONS
19.If x=asin nt +b cos nt,find the acceleration of the particle∈terms t ∧show that
¨x=−n2 x
Solution
˙x=ancos nt−nbsin nt
¨x=−a n2 sin nt−b n2 cos nt
¨x=−n2 (asin nt+bcos nt)
But x=sin nt +b cos nt
Replacing in ¨x therefore becomes
¨x=−n2 x
b).Find the amplitude and the period of the motion
Given that x= asin nt + b cos nt
C2=A2 + B2
Where: A is the amplitude of the first function ( asin nt ¿
B is the amplitude of the second function( b cos nt ¿
and C is the resultant of the amplitudes of the two functions
C2=A2 + B2
To get the resultant of the two amplitudes we take the square roots of both sides of the equation to get
c which is the resultant of the two amplitudes
√ C2=√ A2 +B2
∴ C= √( A)2 +(B)2( the resultant of the two amplitudes)
And now since a and b are the amplitudes of the first and second function respectively
Plugging in becomes as follows;
c= √(a)2 +(b)2
Resultant amplitude= √( a)2 +(b)2
Period is given by

T= 2 π
w but w (angular velocity is gotten from the equation as it is in the form of ASin wt + Bcos wt
Therefore w=n
Meaning T= 2 π
n sec
c).Calculate the maximum velocity
¨x=−n2 (asin nt+bcos nt) is the acceleration function
We then equate to zero to get the maximum velocity
−n2 ( asin nt+bcos nt ) =0
Meaning either −n2 =0∨asinnt +bcos nt=0
n cannot be zero meaning that ( asin nt+bcos nt )=0
further for it to be maximum a=o
asin nt+ bcos nt=0
tan nt=−b
a
t=¿ ¿
˙x=ancos nt−nbsin nt
Replacing in the above equation becomes
˙x=ancos n ¿ ¿
Simplifying it further becomes
˙xmax=ancos ¿ ¿
21. The displacement of a particle is given by x¿ 2 sin 3t−cos 3 t
a) Show that the particle is moving in SHM
˙x=6 cos 3 t +3 sin 3 t on differentiating the displacement of the particle
Which therefore gives the velocity of the particle
Taking the second derivative also becomes ¨x=−18 sin 3 t +9 cos 3 t

This can also be rewritten as ¨x=−9(2 sin3 t−cos 3t )
This therefore means that acceleration is oppositely proportional to the displacement and we can
conclude that it is a Simple Harmonic Motion (SHM)
b) Find the amplitude and the period of the motion
Given A sin wt + Bcos wt
x¿ 2 sin 3t−cos 3 t
The amplitude = √ ( A2+ B2 )
=√ ¿ ¿
=√5
=2.236 m
Given that x¿ 2 sin 3t−cos 3 t which corresponds to X=A sin wt +Bcos wt
W(angular velocity) from the above expression=3
w=2 πf but f = 1
T ;therefore T = 2 π
w
Plugging in w=3 from the expression gives;
T= 2 π
w = 2 π
3
=2.094
W=3 that is why I believe that T ¿ 2 π
3 and not 2 π
√6 as the student says
c) Find the greatest speed that the particle reaches
¨x=−9(2 sin3 t−cos 3t )
For maximum speed acceleration =0
¨x=−18 sin 3 t+ 9 cos 3 t
sin 3 t
cos 3 t = 9
18
3 t=tan−1 0.2
t=0.0658 seconds
The greatest speed that the particle reaches is given by ˙x=6 cos 3 t +3 sin 3 t
Plugging in 0.0658 seconds becomes

˙x=6 cos 3 ( 0.0658 ) + 3 sin3 (0.0658)
˙x=5.883+0.5884
˙x=6.4714 m
s
22. A particle moves in SHM so that
v2=12 x−6 x2
a) Find the equilibrium and endpoints of the motion
v2=12 x−6 x2 ˙x2=12 x−6 x2
To get displacement we integrate the function on both sides which gives
∫ ˙x2 dx=∫12 x −6 x2 dx
x2=6 x2−2 x3
Also the equilibrium position of the motion is when v=0
12 x−6 x2=0
x (12−6 x )=0
The equilibrium position of the motion is equivalent to 2 m
X=2 m and x=0 m
The end points can be gotten when x=o
0=6 x2−2 x3
⇒ 6 x2=2 x3
X=3 and x=0
b) Find the period of the motion
v2=12 x−6 x2
but the endpoint is3 m
Therefore the angular velocity is equivalent to v= √ 12 ( 3 ) −6(32)=− √ 18=-4.24 rad/s

T= 1
f = 2 π
w = 2 π
4.24 =1.48 seconds