Oscillations Test 3

U=Kx³
F=-du/ 
F=-Kx ------(2)
F=-3kx² -----(1)
Comparing equation 1 and 2
K=-3kx
T=2π√m/K
Because, x=Asinwt x∝A
T∝1/√k ∝1/√A

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.

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