# NCERT : Oscillations

Q: 1. Which of the following examples represent periodic motion ?

(a)A swimmer completing one (return) trip from one bank of a river to other bank

(b) A freely suspended bar magnet displaced from its N-S direction and released,

(c) A hydrogen molecule rotating about its centre of mass,

(d) An arrow released from a bow.

Sol:
(a) Not a periodic motion as he does not take a definite time to repeat his motion

(b) Is a periodic motion as it oscillates about N-S direction.

(c) Is a periodic motion.

(d) Not a periodic motion.

Q:2. 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 oscillatory mercury column in a U-tube

(c) Motion of a ball bearing inside a smooth curved path, when released from a point slightly above the lower most position.

(d) General vibrations of a polyatomic molecule about its equilibrium position.

Sol:
(a) It is periodic but not SHM because it is not a to and from motion about a fixed point.

(b) S.H.M.

(c) S.H.M.

(d) It is periodic but not S.H.M.

Q:3. Which of the following functions of time represent
(a) simple harmonic
(b) periodic but not simple harmonic and
(c) non-periodic motion ?
Give period for each case of periodic motion. (ω is any positive constant)
(a) sinωt-cosωt
(b) sin³ωt
(c) 3 cosω(π/4 – 2ωt)
(d) cosωt+cos3ωt+cos5ωt
(e) e(-ω²t²)
(f) 1 + ωt+ ω²t²

Sol:(a) sinωt-cosωt $\displaystyle = \sqrt{2}[\frac{1}{\sqrt{2}} sin\omega t – \frac{1}{\sqrt{2}} cos \omega t ]$

$\displaystyle = \sqrt{2}[ sin\omega t cos \pi/4- cos \omega t sin\pi/4]$

$\displaystyle = \sqrt{2}[ sin (\omega t -\pi/4)]$

It is an SHM and time period is 2π/ω

(e) e(-ω²t²) is an exponential function which does not repeat. Hence , it is non-periodic.

(f) 1 + ωt+ ω²t² is also non periodic.

Q: 4. Which of the following relationships between the acceleration a and displacement x of a particle involve SHM ?
(a) a = 0.7x (b) a = -200x² (c) a = -10x (d) a = 100 x³

Sol:

In SHM, a ∝ – x

Hence , (c) represents SHM.

Q:5.The motion of a particle in S.H.M is described by the displacement function , x = A cos(ωt + φ). If the initial (t = 0) position of the particle is 1 cm and the initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency is π s-1. If the displacement function be x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

Sol:
Here , t = 0 , x = 1 cm, v = ω cm/s , φ = ? ; ω = π/s

x = A cos(ωt + φ)

1 = A cos (π × 0 + φ)

1 = A cosφ  …(1)

Since , x = A cos(ωt + φ)

Differentiating w.r.t time

dx/dt = – A ω sin(ωt + φ)

ω = -A ω sin(π × 0 + φ)

1 = -A sin φ …(ii)

Squaring & adding (i) and (ii)

2 = A2 (cos2φ + sin2φ )

A2 = 2

A = √2 cm

dividing , tanφ  = -1

φ = 3π/4 or , 7π/4

For x = B sin (ωt + α), Solving similarly

We get B = √2 cm and phase angle , α = π/4 or , 5π/4

Q: 6. 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 spring, when displaced and released, oscillates with period of 0.60 s. What is the weight of the body ?

Sol:
m = 50 kg , Max. extension = y = 20 – 0 = 20 cm = 0.2 m, T = 0.6 s

Force = F = mg = 50 × 9.8 N

k = F/y

k = 50 × 9.8/0.2 = 2450 N/m

$\displaystyle T = 2 \pi \sqrt{\frac{m}{k}}$

$\displaystyle m = \frac{T^2 k}{4 \pi^2}$

$\displaystyle m = \frac{(0.6)^2 \times 2450}{4 (3.14)^2}$

m = 22.36 kg

Weight of the body = mg = 22.36 × 9.8 = 219.1 N

Q: 7. A spring of force constant 1200 N/m is mounted on a horizontal table is as shown in the figure. A mass of 3 kg is attached to the free and of the spring, pulled sideways to a distance of 2 cm and released.
(a) What is the frequency of oscillation of the mass?
(b) What is the maximum acceleration of the mass?
(c) What is the maximum speed of the mass?

Sol: k = 1200 N/m; m = 3 kg; a = 2 cm = 0.02 m

(a)Frequency $\displaystyle \nu = \frac{1}{T} = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$

$\displaystyle \nu = \frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}}$

= 3.2 s-1

(b) Acceleration = ω²y = (k/m ) y ;

Max. acceleration = ka/m

=(1200 ×0.02)/3

= 8 m/s²

Max. speed = aω

= a√(k/m)

= 0.02×√(1200/3)

= 0.4 m/s

Q:8. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1 m. If the piston moves with S.H.M with an angular frequency of 200 rev/min., What is its maximum speed ?

Sol:

a = 1/2 m , ω =200 rev/min

vmax = aω = 100 m/min

Q: 9. The acceleration due to gravity on the surface of the motion is 1.7 m/s². 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 = 9.8 m/s²)

Sol:
gm = 1.7 m/s², ge = 9.8 m/s²,

Tm = ?, Te = 3.5 s

$\displaystyle T = 2 \pi \sqrt{\frac{l}{g}}$

$\displaystyle \frac{T_m}{T_e} = \sqrt{\frac{g_e}{g_m}}$

$\displaystyle T_m = T_e \times\sqrt{\frac{g_e}{g_m}}$

$\displaystyle T_m = 3.5 \times\sqrt{\frac{9.8}{1.7}}$

= 8.4 sec

Q:10. 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 1.5s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire.

Sol:
m = 10 kg, R = 15 cm = 0.15 m,

T = 1.5 s , α = ?
Moment of inertia of the disc , I = (1/2) mR²
I = (1/2)×10×(0.15)²

$\displaystyle T = 2\pi \sqrt{\frac{I}{\alpha}}$

$\displaystyle \alpha = \frac{4\pi^2 I}{T^2}$

$\displaystyle \alpha = 4 (22/7)^2 \times \frac{1}{2} \times \frac{10 \times (0.15)^2}{(1.5)^2}$

Q:11. A body describes SHM with an amplitude of 5cm and a period of 0.2s. Find the acceleration and velocity of the body when the displacement is (a) 5cm, (b) 3cm, (c) 0cm
Solun:
a = 5cm = 0.05m, T = 0.2s, ω=2π/T=10π rad/s
(a)When y = 5 cm = 0.05 m
Acceleration $\displaystyle A = – \omega^2 y = -(10\pi)^2 \times 0.05$

A = – 5 π2 m /s2

$\displaystyle v = \omega \sqrt{a^2 -y^2}$

$\displaystyle v = \omega \sqrt{(0.05)^2 -(0.05)^2} = 0$

(b) When y = 3 cm = 0.03 m

Acceleration $\displaystyle A = – \omega^2 y = -(10\pi)^2 \times 0.03$

A = – 3 π2 m /s2

$\displaystyle v = \omega \sqrt{a^2 -y^2}$

$\displaystyle v = 10\pi \sqrt{(0.05)^2 -(0.03)^2} = 0.4\pi m/s$

(c) when y = 0

A = 0
$\displaystyle v = 10\pi \sqrt{(0.05)^2 -0} = 0.5\pi m/s$