Fourier Series: Two Marks Important Questions and Answers

Fourier Series:

2 Marks Important Questions with Answer

 

PART‒A QUESTIONS AND ANSWERS

 

I. Problems under (0, 2π), (0, 2l), (−π, π), (−l, l)

 

1. When does a function possess a Fourier series expansion in terms of trignometric terms?

(or) State the conditions for f(x) to have Fourier series expansion.

(or) Explain Dirichlet's conditions.

Solution:

the function f(x) = sin(1/x) attains its maximum and minimum values when 1/x =

x= 2/ (2n-1)π       …..(1)

where n is zero or an integer.

For large values of n as n→∞, the values of x as given by (2) tend to become indefinitely small and to be crowded near to the value of x = 0.

Hence, the function has an infinite number of maxima and minima near x = 0, so it does not satisfy one of the Dirichlet's conditions. It cannot be expanded in a Fourier series in the range -π ≤ x ≤π in which the point x = 0 is included.

 

2. State whether y = tan x can be expanded as a Fourier series. If so how? If not why?

Solution: tanx cannot be expanded as a Fourier series. Since, tanx does not satisfy Dirichlet's conditions.

 [tanx has infinite number of infinite discontinuities.]

 

3. Find the sum of the Fourier series for

f(x) = x    0 < x < 1

= 2    1 < x < 2

at x = 1.

Solution:

 x = 1 is a finite point of discontinuity (in the middle) of(0, 2)

 [f (1−) ≠ f (1+)]

 

4. If the Fourier series for the function

f(x) = x         0 < x < π

= sin x ; π < x < 2π is

Solution: Put x = π/2 is a point of continuity.

 

5. Find the constant term in the Fourier series corresponding to f(x) = cos² x expressed in the interval (‒π, π).

Solution:

 

6. Write a₀, a_n in the expansion of x + x³ as a Fourier series in (‒ π, π).

Solution:

Let f(x) = x+x³

f (−x) = (‒x)+(‒x)³

= ‒x‒x³ = (x+x³) = ‒ f(x)

Therefore f(x) is an odd function.

Hence a₀=0 and a_n = 0

 

7. What are the constant term a₀ and the coefficient of cos nx, a_n in the Fourier series expansion of f(x) = x − x³ in (−π, π) ?

Solution:

 f(x) = x‒x³

 f(‒x) = (‒x)‒(‒x)³ = −x+x³ = − (x − x³) = −f(x)

 f(x) is an odd function.

Hence, in the Fourier series a₀ = 0 and a_n = 0

 

8. In the Fourier expansion of f(x) = 

in (−π, π), find the value of b_n, the coefficient of sin nx.

Solution:

 Given function is an even function.

Hence, the value of b_n = 0

 

9. If f(x) = x² + x is expressed as a Fourier series in the interval (‒2, 2) to which value this series converges at x = 2?

Solution:

 x=2 is a point of discontinuity in the extremum.

 

10. Find b_n in the expansion of x² as a Fourier Series in (‒ π, π).

Solution:

Given f(x) = x² is an even function in the interval (‒ π, π)

  b_n = 0

 

11. If f(x) is an odd function defined in (‒l, l), what are the values of a₀ and a_n ?

Solution:

Given f(x) is an odd function in the interval (−l, l)

 a₀= 0, a_n = 0

 

12. Find the Fourier constants b_n for x sin x in (‒π, π).

Solution:

Given f(x) = x sinx in (‒π, π)

f(‒x) = (‒x)sin (‒x) = (‒x) [‒ sin x]

= x sin x = f(x)

f(x) is an even function

Hence b_n = 0

 

13. If f(x) =  and f(x) = f(x + 2 π) for all x, find the sum of the Fourier series of f(x) at x = π.

Solution:

Given: f(x) = 

To find f(x) at x = π

 x = π is a discontinuous point in the middle.

 

14. Determine the value of a_n in the Fourier series expansion of f(x)=x³ in ‒π < x < π.

Solution:

Let  f(x) = x³

 f(‒x) = (‒x)³ = ‒x³ = ‒f(x)

Therefore f(x) is an odd function.

Hence a₀ = 0 and a_n = 0

 

15. If f(x) = 2x in the interval (0, 4), then find the value of a₂ in the Fourier series expansion.

Solution: Here 2l = 4; l = 2

 

II. Problems under Half range series

 

1. Find half range sine series for, f(x) = k in 0 < x < π.

Solution: The sine series of f(x) in (0,π) is given by

 

2.(a) Sketch the even and odd extension of the periodic function

f(x) = x² for 0 < x < 2

Solution:

2.(b) Sketch the graph of one even and one odd extension of

f(x)=x³ in [0, 1]

Solution:

 

3. The cosine series for f(x) = x sin x for 0 < x <π is given as

 

4. Expand f(x) = 1 in a sine series in 0 < x <π.

Solution: The sine series of f(x) in (0,π) is given by

 

5. Find the Fourier sine series of f(x) = x in 0 < x < 2.

Solution: In the interval 0 < x < 2 the half range sine series for

 

6. To which value, the half range sine series corresponding to f(x) = x² expressed in the interval (0, 2) converges at x = 2?

Solution: Given f(x) = x²

x = 2 is a finite point of discontinuity and also it is an end point.

Since f(x) = 

the half‒range sine series corresponding to f(x) = x² in the intervel (0, 2) converges at x = 2 is

[f (−2) + f (2)] /2  = [‒4 + 4]/2 = 0

Therefore at x=2, the series converges to 0.

 

III. Problems under Parseval's identity, R.M.S value

 

1. State Parseval's identity for the half‒range cosine expansion of f(x) in (0, 1).

Solution:

 

2. Find the root mean square value of the function f(x) = x in the interval (0, l).

Solution:

 

3. Define root mean square value of a function f(x) in a < x < b.

Solution: Let f(x) be a function defined in an interval (a, b) then

is called the root mean square (or) effective value of f(x) and is denoted by .

 

4. Find the root mean square value of f(x) = x (l‒x) in 0 ≤ x ≤ l.

Solution:

 

5. Define R.M.S value of a function f(x) in c < x <c + 2l.

Solution:

 

6. State TRUE or FALSE: Fourier series of period 20 for the function f(x) = x cos (x) in the interval (‒10, 10) contains only sine terms. Justify your answer.

Solution:

TRUE. f(x) = x cos x is odd in (‒10, 10)

 a₀=0 and a_n = 0

So, we get only sine terms.

 

IV. Problems under Solution of one‒dimensional wave equation and One‒dimensional equation of Heat conduction.

 

1. What conditions are assumed in deriving the one dimensional wave equation?

The wave equation is 

This gives y (x, t), the transverse vibration of a string stretched to a constant tension T. In deriving this equation, we make the following assumptions.

1. The motion takes place entirely in one plane i.e., xy plane.

2. We consider only transverse vibrations, the horizontal displacement of the particles of the string is negligible.

3. The tension T is constant at all times and at all points of the deflected string.

4. T is considered to be so large compared with the weight of the string and hence the force of gravity is negligible.

5. The effect of friction is negligible.

6. The string is perfectly flexible, i.e., it can transmit tension but not bending or shearing forces.

7. The slope of the deflection curve at all points and at all instants is so small that sina can be replaced by a, where a is the inclination of the tangent to the deflection curve,

 

2. What are the various solutions of ?

Thus the various possible solutions of the wave equations are

y(x, t) = (A₁e^Px + A₂e^‒px) (A₃ e^cpt + A₄ e^‒cpt)

y(x, t) = (A₅ cos px + A₆ sin px) (A₇ cos cpt + A₈ sin cpt)

 y(x, t) =  (A₉x+A₁₀) (A₁₁t + A₁₂)

 

3. A string is stretched and fastened to two points l apart. Motion is started by displacing the string into the form y = y₀ sin(πx / l) from which it is released at time t = 0. Formulate this problem as the boundary value problem.

Solution:

The displacement function y (x, t) is the solution of the wave equation.

The boundary and initial conditions are

(i) y (0, t) = 0 for all t≥0

(ii) y (l,t) = 0 for all t≥0

(iii) (∂y/∂t)_(x,0) = 0

(iv) y (x, 0) = f(x) = y₀ sin (πx/l)

 

4. A string of length 2 l stretched to a constant tension T, is fastened at both the ends and hence fixed. The mid point of the string is taken to a height b and then released from rest in that position. It is desired to solve for transverse vibrations of the string. Write the governing equation and the corresponding conditions.

Solution:

The wave equation is ∂²y/∂t² = a² ∂²y/∂x²

From the given problem, we get the following boundary and initial conditions.

(i) y (0, t) = 0 for all t ≥ 0

(ii) y (2l, t) = 0 for all t ≥ 0

(iii) (∂y/∂t)_{(x, 0)} = 0, 0 ≤ x ≤ l

 

5. What is the constant a² in the wave equation

 u_tt = a² u_xx ?

(OR)

In the wave equation  what does c² stand for?

Solution:

 a² (or) c² =  T/m = Tension / mass per unit length of the string

 

6. State the suitable solution of the one dimensional heat equation

Solution: u(x, t) = (A cos px + B sin px) 

 

7. State the governing equation for one dimensional heat equation and necessary conditions to solve the problem.

Solution: The one dimensional heat equation is

where u(x, t) is the temperature at time t at a point distant x from the left end of the rod.

The boundary conditions are

(i) u (0, t) = k₁°C for all t≥0

(ii) u (l,t) = k₂° C for all t≥ 0

 ( l being the length of the one dimensional rod)

The initial condition is

 (iii) u (x, 0) = f(x), 0 < x < l

 

8. Write all variable separable solution of the one dimensional heat equation u_t = a²u_xx.

u(x, t) = (A₁e^λx + B₁e^−λx) C₁ 

u(x, t) = (A₂ cos λx + B₂ sin λx) C₂ 

 u(x, t) = (A₃x + B₃) C₃

 

9. Write down the diffusion problem in one‒dimension as a boundary value problem in two different forms.

Solution:

 one dimensional heat flow.

Here a² = k/ρc is called the diffusivity.

In the steady state ∂²u/∂x² = 0

 

10. State any two laws which are assumed to derive one dimensional heat equation.

Solution:

(i) Heat flows from higher to lower temperature.

(ii) The rate at which heat flows across any area is proportional to the area and to the temperature gradient normal to the curve. This constant of proportionality is known as the thermal conductivity (k) of the material. It is known as Fourier law of heat conduction.

 

11. In steady state conditions, derive the solution of one dimensional heat flow equation.

Solution: The p.d.e. of unsteady one dimensional heat flow is

             ……(2)

In steady state condition, the temperature u depends only on x and not on time t

Hence ∂u/∂x = 0

 (1) reduces to d²u/dx² = 0

             ……(2)

The general solution is u = ax + b where a, b are arbitrary.

 

12. Write the boundary conditions and initial conditions for solving the vibration of string equation, if the string is subjected to initial displacement f(x) and initial velocity g(x).

Solution: The wave equation is 

  (a) y (0, t) = 0

  (b) y (l,t) = 0

  (c) (∂y/∂t)_(x,0) = g(x)

  (d) y(x, 0) = f(x)

 

13. The ends A and B of a rod of length 10 cm have their temperature kept at 20°C and 70°C. Find the steady state temperature distribution on the rod.

Solution: The steady state temperature distribution on the rod

Here,

 a: Temperature at end x = 0, i.e., a = 20°C

 b: Temperature at end x = l, i.e., b = 70°C

 l: Length of the rod, i.e., l = 10 cm

  (1) ⇒ u (x) = ( (70‒20)/10 )x + 20 = 5x + 20

 

14. Explain the various variables dominating the wave equation.

Solution: The wave equation is

y → displacement

x → distance of the point on the string from fixed point along x direction

t → time