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QUESTION: 1

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 x-axis. If the amplitude of oscillation is a, then its time period, T is

Solution:

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

QUESTION: 2

The dimensions and unit of phase constant Φ is

Solution:

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

QUESTION: 3

If the reference particle P moves in uniform circular motion, its projection along a diameter of the circle executes

Solution:

SHM is a 1D projection of 2D UCM.

QUESTION: 4

The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration is

Solution:

QUESTION: 5

If the sign in equation F = -kx is changed what would happen to the motion of the oscillating body?

Solution:

QUESTION: 6

Choose the correct time period of the function sin ωt + cos ωt

Solution:

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

QUESTION: 7

The velocity and acceleration amplitudes of body executing simple harmonic motion is

Solution:

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.

QUESTION: 8

At what distance from the mean position would the K.E of a particle in simple harmonic motion be equal to its potential energy?

Solution:

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

QUESTION: 9

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.

Solution:

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=230-180

P.Σ=50J

QUESTION: 10

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

Solution:

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

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