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Question-1

The soldiers marching on a suspended bridge are advised to go out of step. Why?

Solution:
If the soldiers while crossing a suspended bridge march in step, the frequency of marching steps of soldiers may match with the natural frequency of oscillations of the suspended bridge. In such a situation, resonance will take place, and the amplitude of oscillations of the suspended bridge will increase enormously, which may cause the bridge to collapse. To overcome this situation, the soldiers are advised to march out of step on a suspended bridge.

Question-2

A girl is swinging in a sitting position. How will the period of swing change, if the girl stands up?

Solution:
The time period of a swing is given by

T = 2π

where 'l' is the distance between the center of gravity of the girl and the point of suspension of the swing. When the girl stands up, her center of gravity is raised and 'l' will decrease. Therefore, the period of swing will decrease.

Question-3

The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change, if the water begins to drain out of the hollow sphere?

Solution:
When the sphere is filled with water, its center of gravity is at the center of the sphere. So let its time period be 'T'. As the water begins to drain out, the center of gravity of the sphere will be lowered and hence length 'l' will increase. Therefore, time period will increase. When all the water is drained out, center of gravity again shifts to the center of the sphere and hence the period will be the same (i.e., T) as when the sphere was full of water.

Question-4

A test-tube of cross-sectional area ‘A’ m2 and mass 'm' kg is floated in water by placing m2kg of lead shots in it. Show that it will execute SHM and deduce an expression for the frequency of oscillations.

Solution:
For depression x, the restoring Force F is given by

F= -weight of water displaced

                   = -A x g

                   =-Kx where k = Ag

Therefore the motion of the test-tube is SHM.

=

=

v ==.

Question-5

Two exactly identical pendulums are oscillating with amplitudes of 2 cm and 6 cm. Calculate the ratio of their energies of oscillation.

Solution:
Total energy of the bob of simple pendulum is given by E = ½mw2r2 i.e. Er2

Therefore, ===.

Question-6

A particle executes SHM of amplitude 30 cm and time period 4s. What is the minimum time required for the particle to move from mean position to a point 15 cm?

Solution:
Let 't' be the time taken by the particle to move from mean position to the extreme point i.e. 15 cm from it.

We know y = r sint

Here y = 15 cm, r = 30 cm, T = 4s

15 = 30sin2t

or sin t ==sin

or t =

or t = s.

Question-7

A block rests on a horizontal table, which is executing SHM in the horizontal plane with amplitude ‘A’. What will be the frequency of oscillation, when the block just starts to slip? The Coefficient of friction = μ.

Solution:
Restoring force on the block = mω2 g

Acceleration in the block = μg

Frequency of oscillation

Question-8

The bob of a vibrating simple pendulum is made of ice. How will the period of swing change when the ice starts melting?

Solution:
The period of swing of a simple pendulum will remain unchanged till the location of centre of gravity of the bob left after melting the ice remains at a fixed distance from the point of suspension. If the centre of gravity of ice bob is raised upwards after melting, then the effective length of the pendulum decreases and hence the time period of swing will decrease. If the centre of gravity of ice shifts to the lower side, then the time period of the swing will increase.

Question-9

Two bodies of masses 1 kg and 3 kg respectively are connected rigidly by a vertical spring. The body of mass 3kg rests on a smooth horizontal table. The force constant of the spring is 484 N/m. From the equilibrium positions, the mass of 1 kg is displaced vertically through a distance of 0.02 m and then released. Calculate the following:

(i) the frequency of oscillation of the mass of 1 kg,

(ii) the maximum velocity of this mass,

(iii) its oscillation energy and

(iv) the amplitude of the harmonic oscillations of the reaction of the table on the mass of 3 kg.

Solution:
Force constant of the spring

K=484Nm-1

(i) Frequency of oscillation of the mass of 1kg

V===x22=3.5Hz.

               

(ii) Maximum velocity of the mass of 1kg

            Vmax = a = a
× 2v

                    =0.022 × × 3.5

                    =0.44ms-1.

Question-10

How can glass windows be broken by a distant explosion?

Solution:
A distant explosion sends out sound waves of large amplitude in all directions. When the sound waves fall on the glass windows, they set the glass into forced oscillations. Since glass is brittle, it breaks as soon as it starts oscillations due to forced vibrations.

Question-11

What happens to the period of a simple pendulum if its length is double? What happens if the mass that is suspended is doubled?

Solution:
Time period of a simple pendulum is

When the length is doubled, the time period

T
= υ 2T

Since the time period is independent of the suspended mass, the time period of the pendulum will remain the same, with increase in mass.

Question-12

Show that the velocity of a particle in SHM is ahead of its displacement by π /2 and lagging behind its acceleration by π /2.

Solution:
Displacement of the particle executing SHM is given by

Y = a sin ω (ω t + ϕ )

Velocity of the particle is obtained by differentiating above equation w.r.t. t i.e.,

v = = aω cos (ω t + ϕ )

= vmax cos (ω t + ϕ )

Thus velocity is ahead of displacement by π /2

Differentiating ‘v’  w.r.t. ‘t’ we have,

F =

= - a ω 2 sin (ω t + ϕ )

= a ω 2 sin [(ω t +θ ) + π )

Thus, acceleration is ahead of displacement byπ or half cycle or acceleration is ahead of velocity by π / 2 or 1/4th of a cycle.

Question-13

If y = a sin ω t + b cos ω t, does it represent SHM? What is the amplitude of motion?
What is the length of the path?

Solution:
y = a sin ω t + b cos ω t

a = r cos θ and b = r sin θ

y = r sin (ω t +θ )

a2+ b2 = r2 (sin2 θ + cos2 θ ) = r2 ---(1)

or r= (taking positive sign)

Substituting the value of r in equation (1), we get (i)

y = sin (ω t +θ )

Therefore the given equation represents SHM

Amplitude of motion, r=
Length of the path, 2r = 2

Question-14

At what distance from the mean position is the K.E. in simple harmonic oscillator equal to P. E?

Solution:
When the displacement of a particle executing SHM is y, then its

K. E = m
ω 2 (a2 - y2)

and P. E. = m
ω 2 y2

If K. E. = P.E., then

m
ω 2 (a2 -y2) = mω 2 y2

or 2y2 = a2 or y = a
v2

Question-15

The maximum acceleration of a simple harmonic oscillator is A0 while the maximum velocity is υ 0. What is displacement amplitude?

Solution:
v0 = ω a

A0 = ω 2 a = (ω a) ω = v0ω

or ω = A0/v0 Thus

a = v0/ω = v0/ (A0/v0) v02/A0

Question-16

Two exactly identical pendulums are oscillating with amplitudes of 2 cm and 6 cm. Calculate the ratio of their energies of oscillation?

Solution:
Total energy of the bob of simple pendulum is given by

E = m
ω 2 r2 i.e., E r2

Therefore




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