Simple Harmonic Motion (SHM) Problem Solution for Physics

This document provides a detailed solution to a physics assignment focusing on Simple Harmonic Motion (SHM). The solution covers multiple aspects of SHM, including calculating the amplitude and period of motion, determining maximum velocity, and proving that a given displacement function represents SHM. The assignment also includes finding equilibrium positions, endpoints, and the angular velocity of the motion. The solutions are presented with clear steps and explanations, making it a valuable resource for students studying physics. The assignment covers topics like calculating amplitude, period, and velocity, as well as determining the equilibrium position and endpoints of the motion.

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