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**Q 1. A particle is executing simple harmonic motion (SHM) of amplitude A, along the x axis, about x = 0. When its potential Energy (P.E.) equals kinetic energy (K.E.), the position of the particle will be (2019)(1) A/2(2) A / (2√2)(3) A/√2(4) AAns:** (3)

(1) 0.77

(2) 0.57

(3) 0.37

(4) 0.17

Ans:

Let ω be the frequency of torsional oscillation.

(1) 3.75 rad s

(2) 1.25 rad s

(3) 2.50 rad s

(4) 5.00 rad s

Ans:

*Disputed question – None of the options is correct.

We know that,

B

And according to second law of motion we have

ma = B – mg

⇒ −ma = ρA(h + Δh)g – ρAhg

⇒ −ma = ρAgΔh

Ans:

We have,

a = −ω

(1) 2

(2) 1/3

(3) 1

(4) 3

Ans:

(1) 10

(2) 1 rad/s

(3) 10

(4) 10

Ans:

Angular frequency of pendulum is

[Δg is due to oscillation of support]

(1) K

(2) K

(3) K

(4) K

Ans:

So, the above equation becomes

(1) [θ = angular amplitude]

If length is doubled then KE will be

From Eq. (1) and Eq. (2), we get

Ans:

According to the relation of G gravitational constant and g acceleration due to gravity, we have

Dividing Eq. (2) by Eq. (1), we get

(1) The wave is propagating along the negative x-axis with speed 25 ms

(2) The wave is propagating along the positive x-axis with, speed 100 ms

(3) The wave is propagating along the positive x-axis with speed 25 ms

(4) The wave is propagating along the negative x-axis with speed 100 ms

Ans:

We know that

y = a sin (ωt + kx)

Given, y = 10

On comparing both the equations, we get

a = 10

Thus, wave is propagating along negative x-axis with speed

Ans:

According the relation between Torque and Moment of Inertia

We know that, α = ω

On comparing Eq. (1) and Eq. (2), we get

Ans:

Let velocity of bob at B is u, which is released by conservation of energy at Point A and B

Let velocity of bob after the collision is v.

By conservation of energy at Point B and C

Substituting the value of v and u in Eq. (2), we get

y = 5(sin 3πt+ 3 cos 3πt) cm

The amplitude and time period of the motion are (2019)

Ans:

We have

y = 5(sin 3πt+ 3 cos 3πt)

By squaring and adding Eq. (2) and Eq. (3), we get

Substituting the value of Eq. (2) and Eq. (3) in Eq. (1), we have

y = A cos θ sin 3πt + A sin θ cos 3πt

⇒ y = A sin (3πt + θ) (4)

⇒ y = 10 sin (3πt + θ)

General wave equation is

y = a sin (ωt + kx)

On comparing with Eq. (4), we get

Ans:

**Q 14. An oscillator of mass M is at rest in the equilibrium position in a potential V = 1/2 k(x – X) ^{2}. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is : (M = 10, m = 5, u = 1, k = 1) (2018)(1) 2/3(2) **

**(3) **

**(4) Ans:** (2)

In first collision mu momentum will be imparted to system. In second collision when momentum of (M + m) is in opposite direction mu momentum of particle will make its momentum zero. on 13

⇒

⇒

**(1) **

**(2) **

**(3) **

**(4) Ans:** (1)

a = Asinωt

b = Asin2ωt

c = Asin3ωt

a + c = A[sinωt

(1)

(2)

(3)

(4)

Ans:

**Q 17. A magnetic needle of magnetic moment 6.7 × 10 ^{-2} Am^{2} and moment of inertia 7.5 × 10^{-6} kg m^{2} is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is (2017)** (3)

(1) 6.98 s

(2) 8.76 s

(3) 6.65 s

(4) 8.89 s

Ans:

For 10 oscillations,

t = 10T = 2π × 1.06

= 6.6568 ≈ 6.65 s

(1) 2 × 10

(2) 2 × 10

(3) 3 × 10

(4) 1 × 10

F = yA ∝ Δt

= 2 × 10

= 9.6 × 10^{4} = 1 × 10^{5} N**Q 19. A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s ^{2} and teh car has acceleration 4 m/s^{2}. The car will catch up with the bus after at time of - (2017)** 4

(2) 15 s

Ans:

(1) 2 s

(2) 5 s

(3) 7 s

(4) 3.5 s

Ans:

(1) 0.01 cm

(2) 0.1 cm

(3) 0.5 cm

(4) 1 cm

Ans:

mω

Total initial energy = 1/2 kA

at x = 2A/3, potential energy =

Kinetic energy at

If speed is tripled, new Kinetic energy =

∴ New total energy =

If next amplitude = A'; then

(1) 0.7 Hz

(2) 1.2 Hz

(3) 1.9 Hz

(4) 0.1 Hz

Ans.

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