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QUESTION: 1

One end of a spring having force constant K is fixed to a vertical wall and other to a body of mass m resting on a smooth horizontal surface. There is another wall at a distance x_{0} from the body. The spring is then compressed by 2x_{0} and released. The time taken to strike the wall from its compressed position is

Solution:

Amplitude of motion A = 2x_{0}

Time to cover from extreme position to mean positon

(i.e. from compressed position to norma position) = T/4

y = A sin ωt

∴ So, total time taken to hit the wall.

QUESTION: 2

A particle executing S.H.M. in a straight line has velocities 8, 7, 4 at three points distant one foot from each other. The maximum velocity of the particle will be

Solution:

we know that

further

ω= 3

QUESTION: 3

A particle is vibrating in simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position is its energy half potential and half kinetic?

Solution:

The total energy E of a particle vibrating in S.H.M. is given by

...(1)

The kinetic energy K is given by

where y = displacement of the particle.

Hence the kinetic energy is half of the total energy when the displacement of the particle is Given that a = 4cm.

QUESTION: 4

A particle of mass m is attached to three springs A, b and C of equal force constants k. Fig.

The particle is pushed slightly against the spring C and released the time period of oscillation will be

Solution:

Let x be the displacement of spring C. The displacement of A = x cos45^{0} = xl√2

Similarly, the displacement of B

Now, forces acting on m in the displacement position of m are kx in the direction C,

Net force in direction C

QUESTION: 5

The ends of a rod of length / and mass m are attached to two identical springs as shown in figure. The rod is free to rotate about its centre O. The rod is depressed slightly at end A and released. The time period of the resulting oscillation is

Solution:

Let the rod be depressed by a small amount x. Both the springs are compressed by x. When the rod is released, the restoring torque is given by

where θ is expressed in raidan. Thus

If is the moment of inertia of the rod about O, then

Since the motion is simple harmonic whose angular frequency is given by

Now Therefore we have

QUESTION: 6

In the case of suspensions shown in the given diagrams, the natural frequencies of the three simple harmonic oscillators will be in the ratio of :

Solution:

First two springs are connected parallel hence displacement in both the spring will be equal say x then by hook's law

F = -kx

By Eqs (i) and (ii)

⇒ (since K_{1} = K_{2})

In third case equivalent spring constant becomes k/2

We know frequency ω =

QUESTION: 7

When a harmonic waves is propagating through a medium, the displacement y of a particle of the medium is represented by y = 10 sin 2π/5 (1800 t-x)

The time period will be:

Solution:

The displacement y = 10 sin 2π/5 (1800 t -x)

Comparing this y equation

we get ...(1)

and ..(2)

By Eq. (i),

⇒ F = 3600/10

= 360

⇒ 1/T = 360

⇒ T = 1/360 s

QUESTION: 8

Two particles execute simple harmonic motions of the same amplitude and frequency along the same straight line. They pass one another travelling in opposite directions, whenever their displacement is half of their amplitude. The phase difference between the two is :

Solution:

We know the S.H.M. can be written as y = 1 sin ωt = a sin θ

For first we have,

For second,

a/2= a sinθ_{2} ⇒ θ_{2} = 30^{0}

Phase difference =

QUESTION: 9

A particle is acted upon by two simple harmonic motions,

x = a sin (ωt + φ)

and y = b sin ωt

Which one of the following diagrams correctly gives the resultant path of the particle when

Solution:

We have x = a sin (ωt + φ) ...(i)

x = a sin ωt and φ = 90^{0}

Thus, when φ = 90^{0 }the Eq. (i) becomes

x = a cos ωt

y = b sin ωt

Squaring and adding

QUESTION: 10

Among the following displacement versus time plots, which ones may represent an overdamped oscillator?

Solution:

Displacement x of curve -B is positive at all instance and the system return to is equilibrium position without large change in oscillations. (and returns is slower)

QUESTION: 11

At the instant t = 0, a particle starts moving due to a force F = F_{0} sinωt, where F_{0}, ω are constants. Which one of the following plots gives the relation between the distance(s) covered by the particle and ωt

Solution:

The equation of force,

F =F_{0} sin ωt

where F_{0} → a constant

ω → angular velocity (a constant)

Now by Newton's second law,

Integrating

when

So,

To draw the graph,

(i) when t = 0, s = 0

(ii) initially as t increases from 0, the rate of increase of s is slower.

So, graph is:

QUESTION: 12

A mass M is suspended from a light spring. An additional mass m added displaces the spring further by a distance x. Now the combined mass will oscillate on the spring with a period :

Solution:

and total mass oscillating = M + m

QUESTION: 13

Among the following displacement versus time plots, which ones may represent a damped oscillator?

Solution:

For damped oscillation amplitude of oscillation is proportional to e^{-yt} • Where γ —► damping parameter. And this is exponentially decreases with time.

QUESTION: 14

The instantaneous position x(t) of a small block performing one-dimensional damped oscillation is x(t) = Here, ω id the angular frequency γ the damping coefficient, A the initial amplitude and α the initial phase. If the value of A and α (with n = 0, 1, 2...) are

Solution:

Given

QUESTION: 15

A man stands on a weighing machine placed on a horizontal platform. The machine reads 50kg. By means of a suitable mechanism the platform is made to execute harmonic vibrations up and down with a frequency of 2 vibrations per second. What will be the effect on the reading of the weighing machine? The amplitude of vibration of the platform is 5 cm. Take g = 10ms^{-2}

Solution:

Here, m = 50kg, v = 2s^{-1}

a = 5 cm = 0.05m

Max. acceleration, a_{max} = ω_{2}a

∴ Max. force on the man

Mini. force on the man

*Multiple options can be correct

QUESTION: 16

Suppose a tunnel is dug through the earth from one side to the other side along a diameter. Neglect all the frictional forces and assume that the earth has a uniform density. Then,

G = 6.67 x 10^{-11 }Nm^{2} kg^{-2}; density of earth = 5.51 x 10^{3} kg m^{-3}

Solution:

Let us assume that a tunnel is made along the earth diameter.

Let a particle of mass m is placed at a distance y from the center of earth in that tunnel.

There will be gravitational attraction force experienced by the particle due to mass of earth of radius y.

F = GMmy/R³

Here G = 6.67x10^(-11) Nm²/Kg²

M = Mass of earth = density * volume = d*V = d*4πR³/3

F = 4πGdR³ym/3*R³ = 4πGdym/3

Linear Force F = ma.

Hence a acceleration at a distance y is given by

a = 4πGdy/3

a = Frequency² * distance

Frequency² = w² = 4πGd/3

W = √(4πGd/3)

Time Period = 2π/w

T = 2π√(3/4πGd)

= √(3π/Gd)

= √(3*3.14/5.51*1000*6.67*10^-11)

= 5062 seconds

= 84.4 minutes.

*Multiple options can be correct

QUESTION: 17

A particle-free to move along the X-axis has potential energy given by U(x) = k [1-exp(-x^{2})] for where k is a positive constant of appropriate dimensions. Then which of the following are incorrect?

Solution:

Given that U(x) = k[1-e^{-x2}]

The graph of U(x) is shown in fig.(1)

Now F(x) = -

Thus force is zero only at following three points:

At any point away from origin (excluding ) the particle is not even in equilibrium.

Further, at finite nonzero values of x, the force is directed towards origin.

If displaced a little about origin, the particle will execute s.H.M. Again, Total mechanical energy = K.E. + P.E.

or

The graph of K.E. is also shown in fig. 1 it is obvious from the graph that K.E. is not the minimum at the origin.

*Multiple options can be correct

QUESTION: 18

One end of each of two identical springs, each of force constant 0.5 N/m, are attached on the opposite sides of a wooden block of mass 0.01 kg. The other ends of the springs are connected to separate rigid supports such that the springs are unstretched and are collinear in a horizontal plane. To the wooden piece kept on a smooth horizontal table is now displaced by 0.02m along the line of springs and released. If the speed of the paper is 0.1 m/s, then which of the following are correct ?

Solution:

Here as k = k_{1} +k_{2} = 0.5 + 0.5 = 1N/m and m = 0.01kg,

As the block is displaced by 0.02m, amplitude will be 0.02m. So the equation of motion along the x-axis will be

..(i)

Where φ is phase constant.

But along the y-axis as paper is moving down, the pointer relative to paper is moving up with constant velocity 0.1 m/s, so that

y = vt = 0.1t

The equation of the path will be obtained by eliminating t between Eqns. (1) and (2) i.e.

x= 0.02 sin(100y+

Now x will be maximum when

sin (100y + = max = 1,

which implies

So that

*Multiple options can be correct

QUESTION: 19

Which of the following statements are correct when white light is incident on a zone plate from a point?

Solution:

The radius of zone plate

where b = distance of point from zone plate

⇒ r ∝ √n

focus f

=> focal length =>

Hence, first statements is true.

The highest intensity occurs at a point for which zone plate is produced. => (iv) statement is wrong for a particular wavelength a number of foci are formed, and (iii) statement is wrong.

Thus, statements (i) and (ii) are right.

*Multiple options can be correct

QUESTION: 20

When an unpolarised light beam passes through a double refracting medium, it splits up into two beams called ordinary ray and extraordinary ray. Which of the following statements are correct?

Solution:

When an unpolarised light beam passes through a double refracting medium it splits into two beams viz extraordinary ray which has different velocity in different directions and other is ordinary ray whose velocity in all direction is same. Both these are plane polarised and the plane of polarisations of these are perpendicular to each other. Thus, refractive index of the ordinary ray remains constant while refractive index of the extraordinary ray does not remain constant.

*Answer can only contain numeric values

QUESTION: 21

A horizontal platform executes up and clown S.H.M. about a mean position. Its period is 2π sec. A mass m is resting on the platform. The greatest value of amplitude (in m) so that the mass ‘m' may not leave the platform i s _________ .

Solution:

Here if R = normal reaction then R-mg

= -mg

or R = mg-ma = m(g-a)

Also Max. acceleration a= Aω^{2}

∴ R = mg - mAω^{2}

For object, not to leave the platform R = 0

∴ mg = - mAω^{2}

*Answer can only contain numeric values

QUESTION: 22

A uniform rope of mass 0.1 kg and length 2.45 m hangs from a ceiling. Find the speed (in m/s) of transverse wave in the rope at a point 0.5m distant from the lower end, (g = 9.8 m/s^{2})

Solution:

As the string has mass and it is suspended vertically, tension in it will be different at different points. For a point at a distance x from the free end, tension will be due to the weight of the string below it .So if M is the mass of string of length L, the mass of length x of the string will be (M/L)x.

So,

Here x

*Answer can only contain numeric values

QUESTION: 23

A uniform rope of mass 0.1 kg and length 2.45 m hangs from a ceiling. Calculate the time (in sec) taken by a transverse wave to travel the full length of the rope (g = 9.8 m/s^{2})

Solution:

At point x the wave travels a distance dx in time dt.

[from eq. 1]

i.e [from eq. 2]

Hence L = 2.45 m so

*Answer can only contain numeric values

QUESTION: 24

A source of sound S is moving with a velocity 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency (in Hz) of the source when it is moving away from observer after crossing him ? The velocity of sound in the medium is 350 m/s.

Solution:

When the source is coming to stationary observer

When the source is moving away from stationary observer

*Answer can only contain numeric values

QUESTION: 25

The mass and diameter of a planet are twice those of the earth. The time period (in sec) of a simple pendulum on this planet is _________ , if it is second's pendulum on earth .

Solution:

For seconds pendulum

= T’ = 2√2 s = 2.83 s

*Answer can only contain numeric values

QUESTION: 26

The vibration of a string can be described by the equation y = (0.75 cm) {cos(2π/ 5cm^{-1})x} sin{(60πs^{-1})t}. The speed (cm/s) of the waves travelling in the string is_______

Solution:

Comparing this with the standard equation of standing wave given as

where velocity of wave v = ω / k

We have,

⇒ = 150 cm/s

*Answer can only contain numeric values

QUESTION: 27

An astronaut is approaching the moon. He sends out a radio signal of frequency 5 * 10^{9} Hz and finds out that the frequency shift in echo received is 10^{3} Hz. Find his speed (in m/s) of approach.

Solution:

If the astronaut (source) at speed v is a approaching the moon, the frequency ‘heard’ by the moon will be

...(1)

Now this frequency will be reflected back by the moon (now source ) towards the astronaut which is observer and moving towards the moon . So the astronaut will hear a frequency

...(2)

Substituting the value of f, from Eq (1) in (2)

or i.e

So substituting the given data

*Answer can only contain numeric values

QUESTION: 28

A car has front - and back - directed speakers mounted on its roof, and drives toward you with a speed of 50 ft/s, as shown in figure. If the speakers are driven by a 1000 Hz oscillator, what beat frequency (in Hz) will you hear between the direct sound and the echo off a brick building behind the car ? (Take the speed of sound as 1000 ft/s.).

Solution:

The sound from the back-directed speaker has Doppler frequency

where c and v are the speeds of sound and the car respectively, and v is the frequency of the sound emitted. As the wall is stationary with respect to the obseiver, v_{b} is also the frequency as heard by the latter. The sound from the front-directed speaker has Doppler frequency

Hence the beat frequency is

*Answer can only contain numeric values

QUESTION: 29

The focal length of a piano convex lens, the convex surface of which is silvered, is 0.3 m. If μ of the lens is (7/4), the radius of curvature (in m) of convex surface is____

Solution:

∴ R = 2 x 1.75 x 0.3 = 1.05 m

*Answer can only contain numeric values

QUESTION: 30

A thin-walled cell of transparent material contains a glass convex lens of μ = 1.5 and radii of curvature 10 cms each exactly at its middle. The empty spaces are filled with a liquid of μ = 4/3. Calculate the power (in diopter) o f the system.

Solution:

The power of the lens is given by

= (1.5 — 1) (2 / R) = 1/R

The power of each water lens (plano-concave)

= 1/3 R

The power of combination

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