Test: Dynamics of Simple Harmonic Motion - JEE MCQ

Mathematical constants don't have any dimensions hence, it is dimensionless and phase constant is in angle therefore, the unit is radian.

SHM is a 1D projection of 2D UCM.

Maximum acceleration=ω²r
= (0.45)²x3
=0.60ms^-2

If the sign is changed in F=-kx then the force and hence acceleration will not be opposite to the displacement. Due to this the particle will not oscillate and would accelerate in the direction of displacement. Hence the motion of the body will become linearly accelerated motion.

If the time period of f(x) = T
then time period of f(ax+b) = aT
the time period of sint+cost= 2π
so, time period of sinωt+cosωt = 2π/ω

Maximum velocity: v = ωA, where ω is uniform angular velocity and a is the radius of the circle in which a reference particle executing S.H.M.
Velocity is maximum at mean positions. The maximum value of velocity is called velocity amplitude in SHM.
Acceleration is maximum at extreme position given by A = - ω²A. The maximum value of acceleration is called acceleration amplitude in SHM.

Let say from some distance x, the KE = PE and as total energy must be conserved and TE = -½ kA²
Thus we get 2PE = ½ kA²
Thus we get 2kx² = kA²
We get x = A / √2

K. Σ=1/2 K(A²-x²)
Max of mean position,
K. Σ=1/2 KA²
=1/2 x4x10⁵x(3x10^-2)²
=180J
T.M. Σ=180+P.Σ
230=180+P.Σ
P.Σ=230-180
P.Σ=50J

The relation between angular frequency and displacement is given as
v=ω√A²−x²​
Suppose
x=A sinω t
On differentiating the above equation w.r.t. time we get
dx/dt​=Aωcosωt
The maximum value of velocity will be [{v{\max }} = A\omega \]
The displacement for the time when speed is half the maximum is given as
v=Aω/2
A²ω²=4ω(A²−x²)
By substituting the value in (1) we get the displacement as
x=A√3/2