Oscillations and Waves: Important Solved Problems

ANNA UNIVERSITY SOLVED PROBLEMS

1. A wire of length 1 metre and diameter 1mm is clamped at one of its ends. Calculate the couple required to twist the other end by 90°. Given modulus of rigidity = 298 Gpa

Given data

n = 298 × 10⁹ Pa

θ = 90° = π/2 radians

r = 0.5 × 10^‒3m

L = 1m

Solution

Formula: Twisting Couple C = πnθr⁴ / 2L

C = [π × 298 × 10⁹ × π × (0.5×10^‒3)⁴] / [2×2×1]

C = 0.18363 / 4

C = 4.59 × 10^‒2 NM

Couple required to twist the other end by 90° = 4.59×10^‒2NM.

2. A wire of radius 0.8mm and length 50cm is twisted through an angle of 45° The module of rigidity of the material of the wire is 8×10¹⁰ N/m². Calculate the work done.

Given data

n = 8 × 10¹⁰N/m²

r = 0.8 × 10^‒3m

0 = 45° = π/4 radians;

L = 0.5m

Solution

Formula: Couple per unit twist C = πnr⁴ / 2L

3. A torsion pendulum is made using a steel wire of diameter 0.5mm and a sphere of diameter 3 cm. The rigidity modulus of steel is 80 GPa and density of the material of the sphere is 11300 kg/m³. If the period of oscillation is 2 second, find the length of the wire.

Given data

n= 80 × 10⁹ Pascals

r = 0.25 × 10^‒3m

R = 1.5 × 10^‒2m

ρ = 11300 kg/m³

T = 2 seconds

Solution

Formula: Rigidity Modulus n = 8πΙl / T²r⁴

For sphere the moment of Inertia I = 2/5 MR²

 l = 3.463m

Length of the suspended wire = 3.463m

4. A cylindrical wire of length 1 m and radius 5 mm is rigidly clamped at one end. Calculate the couple required to twist the free and through an angle 45°. The rigidity modulus of the material of the wire is 200×10⁹ Pa.

Given data

L = 1 m

r = 5 × 10^‒3 m

0 = 45° = (45° × π) / 360

n = 200×10⁹ Pa

Solution

Twisting Couple C = πnθr⁴ / 2L

C = (π×200×10⁹ × 45 × 2π × (5×10^‒3)⁴) / (2×1×360)

C = 154.056 Newton meter

Couple per unit twist C = 154.056 Nm

5. An elastic wire is cut into half its original length. How will it affect the maximum load the wire can support?

Solution:

We know

(i) E = (Load/Area) × (original length/change in length)

(ii) Tensile strength = Maximum tensile load / Original cross sectional area

When Original Length is halved, strain is changed so that the maximum load that the wire can support will remain the same as Elastic modulus is constant.

6) The classroom door is of width 50cm. If the handle of the door is 20 cm from the edge and the force of 5 N is applied on the handle, compute the torque.

Solution

Given data

Width of the door = 50 × 10^‒2 m

Handle of door located = 20 × 10^‒2 m

Line of action = [ 20×10^‒2 / 2 ] = 10×10^‒2m

Force aplied (F) = 5N

Torque = ?

Formula

Torque (τ) = Force applied (F) × Level Arm distance (d)

Here Lever arm distance d = Width of the door ‒ Line of action

(i.e.) d = [ 50 × 10^‒2 ] ‒ [  20×10^‒2 / 2 ]

(or) d = [ 50×10^‒2 ] ‒ [ 10×10^‒2 ]

(or) d = 40 × 10^‒2

Torque (τ) = 5 × 40 × 10^‒2

(or) (τ) = 200 × 10⁻²

 (τ) = 2 NM

∴ Torque (τ) = 2 NM

ADDITIONAL SOLVED PROBLEMS

1. A particle executing a S.H.M of period 2π seconds has a total energy of 10.24 × 10^‒4 Joule. If the displacement of the particle at π/4 second is 0.08√2 metres, then find the amplitude of the particle.

Solution

Given Data

E = 10.24 x 10^‒4J;

T = 2π sec and

y = 0.08 √2 metres;

t = π/4 sec.

Formula:

In S.H.M. The displacement of a particle is given by

y = A sin ωt

y = A sin (2π/T)t

0.08√2 = Asin [2π/2π . π/4]

0.113=A sin π/4

0.113 = A/√2

A = 0.113× √2

A = 0.1598 m

Amplitude of the particle = 0.1598 m

2. Calculate the maximum amplitude of velocity, for a particle executing S.H.M of period 10sec and amplitude 5.0 cm.

Solution

Given Data

Displacement amplitude (A) = 0.05 m

T= 10 sec.

Formula:

The maximum amplitude of velocity is

Aω = 0.05 × 2π/T

Aω = (0.05×2×3.14) / 10

Aω = 0.0314 m/sec

The Maximum Amplitude of velocity = 0.0314m/s

3. Find the natural frequency for a material of mass 1 kg attached to a spring of stiffness constant 16 N/m.

Solution

Given Data

Force constant k = 16 N/m and

m = 1 kg.

Formula:

For natural frequency

n = 1/2π . √(k/m)

n = 1/2π . √(16/1)

n = 2/π

n = 2 / 3.14

n = 0.64Hz

Natural frequency n=0.64 Hertz

4. A simple pendulum of one meter length is hang at one end. Considering the oscillations to be of small displacements, find the period of oscillation, if the mass of pendulum is 2 kg. (Given g = 9.8 m/s²).

Solution

Given Data

Displacement = 1 m

mass (m) = 2 kg

Formula:

Time period T = 2π √[l/g]

 T=2×3.14×√[1/9.8]

T=6.28×0.319

∴ T = 2.003 sec.

Period of oscillation = 2 seconds

5. A lift is ascending at an acceleration of 3 m/sec². What is the period of oscillation of simple Pendulum of length one meter suspended in the lift? [Given: Acceleration due to gravity = 9.8 m/g²].

Solution

Given Data

Acceleration of lift = 3 m/sec²

length l =1 metre.

The lift is ascending at an acceleration 3 m/sec² and acceleration due to gravity is g = 9.8 m/sec².

Hence the total acceleration (g') is 9.8 + 3 = 12.8 m/sec²

Formula:

Time period T = 2π√[l/g']

T=2π √(1/12.8)

T = 2×3.14×0.2795

T=1.755 sec

or T=1.76

Period of Oscillation = 1.76 seconds

6) A uniform metal disc of diameter 0.1 m and mass 1.2kg is fixed symmetrically to the lower end of a torsion wire of length 1m and diameter 1.44×10^‒3m, the upper end is fixed. The time period of torsional oscillations is 1.98s. calculate the modulus of rigidity of the material of the wire.

Solution

Rigidity modulus of the material = 3.578×10¹⁰ Nm^‒2

ASSIGNMENT PROBLEMS

1. A square metal bar of 25mm side, 0.38mm long and mass 0.8 kg is suspended by a wire of 0.4 mm long and 0.5 mm radius. It is observed to make 50 oscillations in 335.7 seconds. Calculate the Rigidity Modulus of the material of the wire. (Ans.34.5 GPa)

2. A wire of 3m long and 3× 10^‒4m in diameter elongates to 1.32×10^‒3m, when stretched by a force of 0.6 kg weight. Find Young's modulus of the material of the wire. (Ans: 1.9×10¹¹Nm^‒2)

3. A wire of length 1m and diameter 10^‒2m is fixed at one end and twisted at the other end through an angle of 70° by applying a couple of value 0.01Nm. Evaluate the rigidity modulus of the wire. (Ans: 8.281×10¹⁰Nm^‒2)

4. A circular disc of mass 0.25 kg and radius 2.5×10^‒2m is fixed perpendicularly to the end of a wire of length 0.3m and diameter 10^‒3m. Find the rigidity modulus of the wire if the period of oscillation is 1.05 sec. (Ans: 3.418×10¹⁰Nm^‒2)