
Q1 Which of the following examples represent periodic motion? (a) A swimmer completing one (return) trip from one bank of a river to the other and back. (b) A freely suspended bar magnet displaced from its NS direction and released. (c) A hydrogen molecule rotating about its center of mass. (d) An arrow released from a bow. Ans: (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 freelysuspended magnet, if displaced from its NS 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.
Q2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation of earth about its axis. (b) motion of an oscillating mercury column in a Utube. (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. (d) general vibrations of a polyatomic molecule about its equilibrium position. Ans: (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 Utube 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.
Q3 Figure 14.27 depicts four xt plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)? Ans: (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.
Q4 Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) nonperiodic motion? Give period for each case of periodic motion (ω is any positive constant): (a) sin ωt  cos wt (b) sin3 ωt (c) 3 cos (π/4  2ωt) (d) cos ωt + cos 3ωt + cos 5ωt (e) exp (ω2t2) Ans: (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:
sin^{3}ωt = 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 + ω^{2} t^{2} is non periodic.
Q5 A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (a) at the end A, (b) at the end B, (c) at the midpoint of AB going towards A, (d) at 2 cm away from B going towards A, (e) at 3 cm away from A going towards B, and (f) at 4 cm away from B going towards A. Ans: (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).
Q6 Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? (a) a = 0.7x (b) a = 200x2 (c) a = 10x (d) a = 100x3 Ans: (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.
Q7 The motion of a body in simple harmonic motion is given by the displacement function, x (t) = A cos (ωt + φ). Given that at t = 0, the initial velocity of the body is ω cm/s and its initial position is 1 cm, calculate its initial phase angle and amplitude? If in place of the cosine function, a sine function is used to represent the simple harmonic motion: x = B sin (ωt + α), calculate the body’s amplitude and initial phase considering the initial conditions given above. [Angular frequency of the particle is π/ s] Ans: 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:
A^{2}(sin^{2} Φ + cos^{2} 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:
B^{2}[ sin^{2} α + cos^{2} 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, ......
Q8 A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body? Ans: 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/s^{2}
F = 50 × 9.8 = 490
∴Spring constant, k = F / l = 490 / 0.2 = 2450 Nm1
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.
Q9 A spring having with a spring constant 1200 N m1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass. Ans: 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) = ω^{2} 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, V_{MAX }= ω A
= 0.02 X √1200/3
Therefore, V_{MAX }= 0.4 m / s
Hence, the maximum velocity of the mass is 0.4 m/s.
Q10 In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of xaxis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase? Ans: 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
Q11 Figures 14.29 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anticlockwise) are indicated on each figure. Obtain the corresponding simple harmonic motions of the xprojection of the radius vector of the revolving particle P, in each case. Ans: ( 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 xaxis, i.e., Phase angel Φ = + π/2
Therfore, the equation of simple harmonic motion for the xprojection 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 xaxis, in the anticlockwise direction. Thus, phase angle, Φ = + π
Therefore, the equation of simple harmonic motion for the xprojection of OB, at time t is:
x = A cos [ 2πt/T + Φ ]
= 2 cos [ 2πt/8 + π ]
= 2 cos ( πt/4 )
Q14 The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed? Ans: Angular frequency of the piston, ω = 200 rad/ min.
Stroke = 1.0 m
Amplitude, A = 1.0 / 2 = 0.5m
The maximum speed (v_{max}) of the piston is give by the relation:
v_{max} = Aω
= 200 x 0.5 = 100 m/min
Q15 The acceleration due to gravity on the surface of moon is 1.7 ms2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms2) Ans: 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 = T^{2} 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.
Q16 ( i ) The time period of a body having simple harmonic motion depends on the mass m of the body and the force constant k: T =2π √m/k A simple pendulum exhibits simple harmonic motion. Then why does the time period of a pendulum not depend upon its mass? ( ii ) For small angle oscillations, a simple pendulum exhibits simple harmonic motion ( more or less). For larger angles of oscillation, detailed analysis show that T is greater than 2π√ l/g. Explain. ( iii ) A boy with a wristwatch on his hand jumps from a helicopter. Will the wrist watch give the correct time during free fall? ( iv ) Find the frequency of oscillation of a simple pendulum that is free falling from a tall bridge. Ans: ( 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.
Q17 A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period? Ans: 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 = v^{2} / R
Where,
v is the uniform speed of the car
R is the radius of the track Effective acceleration ( a_{aeff} ) is given as :
a_{aeff} = √ g^{2} +(v^{2}/R)^{2}
Time period, T = 2π √ l / a_{aeff}
Where, l is the length of the pendulum
Therefore, Time period T
Q23 A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = α ø, where J is the restoring couple and ø¸ the angle of twist). Ans: 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 mr^{2}
= 1/2 x 10 x (0.15)^{2}
= 0.1125 kg m^{2}
Time period, T = 2π underroot 1 / α
α is the torsional constant.
α = 4 π^{2} l / T^{2}
= 4 x π^{2 }x 0.1125 / (1.5)^{2}
= 1.972 Nm/rad
Hence, the torsional spring constant of the wire is 1.972 Nm rad^{1}.