Welcome to the Chapter 14 - Oscillations, Class 11 Physics NCERT Solutions page. Here, we provide detailed question answers for Chapter 14 - Oscillations. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.
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(b) and (c)
(a) The swimmer’s motion is not periodic. The motion of the swimmer between the banks of a river is back and forth. However, it does not have a definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same.
(b) The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a definite period of time.
(c) When a hydrogen molecule rotates about its centre of mass, it comes to the same position again and again after an equal interval of time. Such motion is periodic.
(d) An arrow released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.
Given,
Spring constant, k = 1200 N/m
Mass, m = 6 kg
Displacement, A = 4.0 cm = 0.04 cm
ω = 14.14 s-1
( i )
Since time is measured from mean position,
x = A sin ω t
x = 4 sin 14.14t
( ii ) At the maximum stretched position, the mass has an initial phase of π/2 rad.
Then, x = A sin( ωt + π/2 ) = A cos ωt
= 4 cos 14.14t
( iii ) At the maximum compressed position, the mass is at its leftmost position with an initial phase of 3π/2 rad.
Then, x = A sin( ωt + 3π/2 )
= -4 cos14.14 t
( 1 ) For time period, T = 4 s
Amplitude, A = 3 cm
At time, t = 0, the radius vector OB makes an angle π/2 with the positive x-axis, i.e., Phase angel Φ = + π/2
Therfore, the equation of simple harmonic motion for the x-projection of OB, at time t is:
x = A cos [ 2πt/T + Φ ]
= 3 cos [2πt/4 + π/2 ]
= -3sin ( πt/2 )
= -3sin ( πt/2 ) cm
( 2 ) Time period, T = 8 s
Amplitude, A = 2 m
At time t = 0, OB makes an angle π with the x-axis, in the anticlockwise direction. Thus, phase angle, Φ = + π
Therefore, the equation of simple harmonic motion for the x-projection of OB, at time t is:
x = A cos [ 2πt/T + Φ ]
= 2 cos [ 2πt/8 + π ]
= -2 cos ( πt/4 )
Angular frequency of the piston, ω = 200 rad/ min.
Stroke = 1.0 m
Amplitude, A = 1.0 / 2 = 0.5m
The maximum speed (vmax) of the piston is give by the relation:
vmax = Aω
= 200 x 0.5 = 100 m/min
Given, Acceleration due to gravity on the surface of moon, g’ = 1.7 m s-2
Acceleration due to gravity on the surface of earth, g = 9.8 m s-2
Time period of a simple pendulum on earth, T = 3.5 s
We know, T=2π √l/g
Where,
l is the length of the pendulum
∴ l = T2 x g / (2π)2
= (3.5)2 x 9.8m / 4 x (3.14)2
The length of the pendulum remains constant.
On moon's surface, time period,
= T '= 2π √l/g'
= 2π √ [ (3.5)2 x 9.8m / 4 x (3.14)2] / 1.7
= 8.4 s
Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.
( i ) The time period of a simple pendulum, T =2π √m/k For a simple pendulum, k is expressed in terms of mass m, as:
k ∝ m
m/k = constant
Thus, the time period T, of a simple pendulum is independent of its mass.
( ii ) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is:
F = –mg sinθ
Where, F = Restoring force
m = Mass of the bob
g = Acceleration due to gravity
θ = Angle of displacement
For small θ, sin θ ∼ θ
For large θ, sin θ is greater than θ. This decreases the effective value of g.
Thus, the time period increases as: T = 2π√ l/g'
( iii ) As the working of a wrist watch does not depend upon the acceleration due to gravity, the time shown by it will be correct during free fall.
( iv ) As acceleration due to gravity is zero during free fall, the frequency of oscillation will also be zero.
The bob of the simple pendulum will experience centripetal acceleration because by the circular motion of the car and the acceleration due to gravity.
Acceleration due to gravity = g
Centripetal acceleration = v2 / R
Where,
v is the uniform speed of the car
R is the radius of the track Effective acceleration ( aaeff ) is given as :
aaeff = √ g2 +(v2/R)2
Time period, T = 2π √ l / aaeff
Where, l is the length of the pendulum
Therefore, Time period T 
(b) and (c) are SHMs
(a) and (d) are periodic, but not SHMs
(a) During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis.
(b) An oscillating mercury column in a U-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.
(c) The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.
(d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.
Mass of the circular disc, m = 10 kg
Radius of the disc, r = 15 cm = 0.15 m
The torsional oscillations of the disc has a time period, T = 1.5 s
The moment of inertia of the disc is:
l = 1/2 mr2
= 1/2 x 10 x (0.15)2
= 0.1125 kg m2
Time period, T = 2π underroot 1 / α
α is the torsional constant.
α = 4 π2 l / T2
= 4 x π2 x 0.1125 / (1.5)2
= 1.972 Nm/rad
Hence, the torsional spring constant of the wire is 1.972 Nm rad-1.
(b) and (d) are periodic
(a) It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case.
(b) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.
(c) It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals of time.
(d) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.
(a) SHM
The given function is:
sin ωt - cos wt
= underroot 2 [1/underroot 2 sin ωt - 1/underroot 2 cos ωt]
= underroot 2 [ sin ωt x cos π/4 - cos ωtx cos π/4]
= underroot 2 ( ωt - π/4)
This function represents SHM as it can be written in the form:
a sin ( ωt + ø )
Its period is: 2π/ω
(b) Periodic, but not SHM
The given function is:
sin3ωt = ¼ [3sin ωt –sin 3ωt]
Even though the two sin ωt represent simple harmonic motions respectively, but they are periodic because superposition of two SIMPLE HARMONIC MOTION is not simple harmonic.
(c) 3 cos (π/4 – 2ωt) = 3 cos (2ωt – π/4)
As it can be written as : a sin ( ωt + Φ) , it represents SIMPLE HARMONIC MOTION
Its period is : π/ω
(d) In cos ωt + cos 3ωt + cos 5ωt, each cosine function represents SIMPLE HARMONIC MOTION, but the super position of SIMPLE HARMONIC MOTION gives periodic.
(e) As it is an exponential function, it is non periodic as it does not repeat itself.
( f ) 1 + ωt + ω2 t2 is non periodic.
(a) Zero, Positive, Positive
(b) Zero, Negative, Negative
(c) Negative, Zero, Zero
(d) Negative, Negative, Negative
(e) Zero, Positive, Positive
(f) Negative, Negative, Negative
Explanation:
The given situation is shown in the following figure. Points A and B are the two end points, with AB = 10 cm. O is the midpoint of the path.
A ————O————B
A particle is in linear simple harmonic motion between the end points
(a) At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this point. Its acceleration is positive as it is directed along AO. Force is also positive in this case as the particle is directed rightward.
(b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point. Its acceleration is negative as it is directed along B. Force is also negative in this case as the particle is directed leftward.
(c)
2cm
←
A.————.——.O——————.B
The particle is executing a simple harmonic motion. O is the mean position of the particle. Its velocity at the mean position O is the maximum. The value for velocity is negative as the particle is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position.
(d)
2cm
←
A.————O——————.——..B
The particle is moving toward point O from the end B. This direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, the particle's velocity and acceleration, and the force on it are all negative.
(e)
3cm
→
A.————.D————.O—————.B
The particle is moving toward point O from the end A. This direction of motion is from A to B, which is the conventional positive direction. Hence, the values for velocity, acceleration, and force are all positive.
(f)
4cm
←
A.————.O————.E——————.B
This case is similar to the one given in (d).
(c) A motion represents simple harmonic motion if it is governed by the force law:
F = -kx
ma = -k
∴ a = -kx / m
Where, F is the force
m is the mass (a constant for a body)
x is the displacement
a is the acceleration
k is a constant
Among the given equations, only equation a = -10
x is written in the above form with k/m = 10.
Hence, this relation represents SHM.
Given,
Initially, at t = 0:
Displacement, x = 1 cm
Initial velocity, v = ω cm/sec.
Angular frequency, ω = π rad/s
It is given that:
x(t) = A cos( ωt + Φ) . . . . . . . . . . . . . . . . ( i )
1 = A cos( ω x 0 + Φ) = Acos Φ
A cos Φ = 1 . . . . . . . . . . . . . . . . ( ii )
Velocity, v = dx / dt
differentiating equation ( i ) w.r.t ‘t’
v = – Aωsin ( ωt + Φ)
Now at t = 0; v = ω and
=> ω = – Aωsin ( ωt + Φ)
1 = – A sin( ω x 0 + Φ) = -Asin(Φ)
Asin(Φ) = – 1 . . . . . . . . . . . . . . . . . . . . . . . ( iii )
Adding and squaring equations ( ii ) and ( iii ), we get:
A2(sin2 Φ + cos2 Φ) = 1 +1
thus, A =√2
Dividing equation ( iii ) by ( ii ), we get :
tan Φ = -1
Thus, Φ =3π/4 , 7π/4
Now if simple harmonic motion is given as :
x = B sin( ωt + α)
Putting the given values in the equation , we get :
1 = B sin ( ω x 0 + α)
Bsin α = 1 . . . . . . . . . . . . . . . . . . . . ( iv )
Also, velocity ( v ) = ω Bcos (ωt + α)
Substituting the values we get :
π = π B sin α
B sin α = 1 . . . . . . . . . . . . . . . . . . . . ( v )
Adding and squaring equations ( iv ) and ( v ), we get:
B2[ sin2 α + cos2 α] =2
Therefore, B = √ 2
Dividing equation ( iv ) by equation ( v ), we get :
B sin α / B cos α = 1
tan α =1 = tan (π/4)
Therefore, α = π/4 , 5π/4, ......
Maximum mass that the scale can read, M = 50 kg
Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m
Time period, T = 0.6 s
Maximum force exerted on the spring, F = Mg
Where,
g = acceleration due to gravity = 9.8 m/s2
F = 50 × 9.8 = 490
∴Spring constant, k = F / l = 490 / 0.2 = 2450 Nm-1
Mass m, is suspended from the balance.
Time period, T = 2π underroot m/k
∴ m = (T / 2π)2 x k
= (0.6 / 2x3.14)2 x 2450 = 22.36 kg
∴Weight of the body = mg = 22.36 × 9.8 = 219.167 N
Hence, the weight of the body is about 219 N.
Given, Spring constant, k = 1200 N/m
Mass, m = 6 kg
Displacement, A = 4.0 cm = 0.04 cm
( i ) Oscillation frequency v = 1/T = 1/2π √k/m
Where,T = time period.
Therefore v =1/2×3.14 √1200/3 = 3.18m/s
Hence, the frequency of oscillations is 3.18 cycles per second.
( ii ) Maximum acceleration (a) = ω2 A
Where, ω = Angular frequency = √k/m
A = maximum displacement
Therefore, a = A( k/m )
a = 0.02 x (1200/3 ) = 8 m /s-2
( iii ) Maximum velocity, VMAX = ω A
= 0.02 X √1200/3
Therefore, VMAX = 0.4 m / s
Hence, the maximum velocity of the mass is 0.4 m/s.
