1 Crore+ students have signed up on EduRev. Have you? 
Potential energy of a particle with mass m is U=k[x]^{3} , where k is a positive constant. The particle is oscillating about the origin on xaxis. If the amplitude of oscillation is a, then its time period, T is
U=Kx^{3}
F=du/
F=Kx (2)
F=3kx^{2} (1)
Comparing equation 1 and 2
K=3kx
T=2π√m/K
Because, x=Asinwt x∝A
T∝1/√k ∝1/√A
The dimensions and unit of phase constant Φ is
Mathematical constants don't have any dimensions hence, it is dimensionless and phase constant is in angle therefore, the unit is radian.
If the reference particle P moves in uniform circular motion, its projection along a diameter of the circle executes
SHM is a 1D projection of 2D UCM.
The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration is
If the sign in equation F = kx is changed what would happen to the motion of the oscillating body?
Choose the correct time period of the function sin ωt + cos ωt
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
The velocity and acceleration amplitudes of body executing simple harmonic motion is
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 =  ω^{2}A. The maximum value of acceleration is called acceleration amplitude in SHM.
At what distance from the mean position would the K.E of a particle in simple harmonic motion be equal to its potential energy?
Let say from some distance x, the KE = PE and as total energy must be conserved and TE = ½ kA^{2}
Thus we get 2PE = ½ kA^{2}
Thus we get 2kx^{2} = kA^{2}
We get x = A / √2
What is the maximum Kinetic energy and minimum potential energy of a harmonic oscillator with amplitude 0.03m, force constant 4×10^{5} N/m and total mechanical energy of 230 J.
K. Σ=1/2 K(A^{2}x^{2})
Max of mean position,
K. Σ=1/2 KA^{2}
=1/2 x4x10^{5}x(3x10^{2})^{2}
=180J
T.M. Σ=180+P.Σ
230=180+P.Σ
P.Σ=230180
P.Σ=50J
A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is
The relation between angular frequency and displacement is given as
v=ω√A^{2}−x^{2}
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
A^{2}ω^{2}=4ω(A^{2}−x^{2})
By substituting the value in (1) we get the displacement as
x=A√3/2
127 videos464 docs210 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
127 videos464 docs210 tests









