# Problems on Simple Harmonic Progressive waves

 Science > Wave Motion > You are Here

#### Example – 1:

• The equation of simple harmonic progressive wave from a source is y =15 sin 100πt. Find the equation of the wave generated if it propagates along the + X-axis with a velocity of 300 m/s.
• Solution:
• Given: Equation of source y =15 sin 100πt, Direction = + X-axis, Velocity of wave v = 300 m/s.
• To Find: Equation of the wave =?

Equation of Source wave is y =15 sin 100πt

Comparing with y =15 sin 2πnt

a = 15 m and 2πn = 100 π

∴  n = 50 Hz

Now, v = nλ

∴  λ = v/n = 300/50 = 6 m

The equation of simple harmonic progressive awve is given by

Hence the equation of S.H.wave is

#### Example – 2:

• The equation of transverse simple harmonic progressive wave is y = 3 sin2π(t/0.04 – x/40), where the length is expressed in cm and the time in seconds. Calculate the wavelength, frequency, amplitude and the speed of the wave.
• Solution:
• Given: Equation of the wave y =3 sin2π(t/0.04 – x/40) cm
• To Find: Wavelength = λ = ?, Frequency = n =?, Amplitude = a =?, speed of wave = v =?

Equation of given wave is

Comparing with

Amplitude = a = 3 cm, Period = T = 0.04 s, Wavelength = λ = 40 cm

We have n = 1/T = 1/0.04 = 25 Hz

Now, v = n λ = 25 × 40 = 1000 cm/s

Ans: Wavelength = 40 cm, Frequency = 25 Hz, Amplitude = 3 cm, speed of wave = 1000 cm/s = 10 m/s

#### Example – 3:

• The equation of a certain sound wave (simple harmonic progressive wave) is given by y = 0.05 sin 10π(t/0.025 – x/8.5). where x and y are in meters and t is in seconds. What are the (1) amplitude (2) frequency (3) wavelength of the wave? What is the velocity and direction of propagation of the wave?
• Solution:
• Given: Equation of the wave y = 0.05 sin 10π(t/0.025 – x/8.5) m
• To Find: Amplitude = a =?, Frequency = n =?,  Wavelength = λ = ?,  speed of wave = v =? and direction = ?

Equation of given wave is

Comparing with

Amplitude = a = 0.05 m, Period = T = 0.005 s, Wavelength = λ = 1.7 m

We have n = 1/T = 1/0.005 = 200 Hz

Now, v = n λ = 200 × 1.7 = 340 m/s

Ther term – x/1.7 shows that the wave is moving in positive direction of x-axis

Ans: Amplitude =0.05 m, Frequency  = 20 Hz,  Wavelength = 1.7 m,

speed of wave = v = 340 m/s and direction = in positive direction of x-axis

#### Example-4:

• The equation of a wave can be represented by y = 0.02 sin 2π /0.5 (320t – x) where x and y are in metres and t is in seconds. Find the amplitude, frequency, wavelength, and velocity of propagation of the wave.
• Solution:
• Given: Equation of the wave y = y = 0.02 sin 2π /0.5 (320t – x) m
• To Find: Amplitude = a =?, Frequency = n =?,  Wavelength = λ = ?,  velocity of wave = v =?

Comparing with

Amplitude = a = 0.02 m, Frequency = n = 640 Hz, Wavelength = λ = 0.5 m

Now, v = n λ = 640 × 0.5 = 320 m/s

Ans: Amplitude =0.02 m, Frequency =640 Hz,  Wavelength = 0.5 m,  velocity of wave = 320 m/s

#### Example – 5:

• Write down the equation of a transverse wave travelling along a stretched string. Given : amplitude 3 m, wavelength = 40 m and frequency = 25 Hz.
• Solution:
• Given: Amplitude = a = 3 m, Vavelength = λ = 40m, Frequency n = 25 Hz
• To Find: Equation of the wave = ?

The equation of progressive wave is in the form

#### Example – 6:

• Write down the equation of a wave moving in the positive direction of x-axis and of amplitude 0.05 m and period 0.04 s traveling along a stretched string with a velocity 12.5 m/s.
• Solution:
• Given: Amplitude = a = 0.05 m, Period = T = 0.04 s, Velocity = v = 12.5 m/s, Direction = + X-axis.
• To Find: Equation of the wave = ?

We have v = n λ

∴  λ = v/n  = v T = 12.5 × 0.04 = 0.5 m

The equation of progressive wave is in the form

#### Example – 7:

• A simple harmonic progressive wave of amplitude 5 cm and frequency 5 Hz is traveling along the positive X-direction with a speed of 40 cm/s. Calculate (1) the displacement at x = 38cm and t = 1 second. (2) The phase difference between two points in the path of the wave separated by a distance of 0.8 cm. (3) The phase difference between two positions of a particle at an interval of 0.01 s.
• Solution:
• Given: Amplitude = a = 5 cm , Frequency  = n = 5 Hz, Velocity = v = 40 cm/s, Direction = + X-axis.
• To Find: 1) Displace ment = y =? when x = 38cm and t = 1 second. 2) ∅ = ? when x = 0.8 cm. 3) ∅ = ? when t = o.o1 s.

We have v = n λ

∴  λ = v/n  =40/5 = 8m

The equation of progressive wave is in the form

• 1) Displace ment = y =? when x = 38cm and t = 1 second.

• ∅ = ? when x = 0.8 cm.

• ∅ = ? when t = o.o1 s.

#### Example – 8:

• The equation of progressive wave is y= 0.01 sin 2π (2t – 0.01x) when all quantities are expressed in SI units. Calculate (a) frequency of the wave. (b) the phase difference between two positions of the same particle at a time interval of 0.25 s. (c) phase difference at a given instant of time between two particles 50 m apart.
• Solution:
• Given: Equation of wave y= 0.01 sin 2π (2t – 0.01x) m
• To Find: 1) Frequency = n =?  2) ∅ = ? when t = o.25 s. 2) ∅ = ? when x = 50 m.

Equation of wave y= 0.01 sin 2π (2t – 0.01x) m

Comparing with

Amplitude = a = 0.01 m, Frequency = n = 2 Hz, Wavelength = λ = 100 m

• ∅ = ? when t = o.25 s.

•  ∅ = ? when x = 50 m.

#### Example – 9:

• The equation of a simple harmonic progressive wave is given by y= 0.002 sin 2π(5t – x/12) where all the quantities are in S.I. units. Calculate the displacement of the particle at a distance of 5 m from the origin after 0.2 s.
• Solution:
• Given: Equation of wave y= 0.01 sin 2π (2t – 0.01x) m
• To Find: 1) Frequency = n =?  2) ∅ = ? when t = o.25 s. 2) ∅ = ? when x = 50 m.

Equation of wave y= 0.002 sin 2π(5t – x/12) m

∴   y = 0.002 sin 2π(5(0.2) – 5/12)

∴   y = 0.002 sin 2π(1 – 5/12)

∴   y = 0.002 sin 2π(7/12)

∴   y = 0.002 sin 7π/6

∴   y = 0.002 sin (6π + π)/6

∴   y = 0.002 sin (π + π/6)

∴   y = – 0.002 sin (π/6)

∴   y = – 0.002 × ½ = – 0.001 m

#### Example – 10:

• The equation of simple harmonic progressive wave of a source is y = 6 sin300πt cm. Write down the equation of the wave. Find the displacement, velocity and acceleration of a point 1.5 m from the source at the instant t = 0.01 s after the start of oscillations. The velocity of propagation of waves is 300 m/s.
• Solution:
• Given: Equation of source is y = 6 sin300πt cm, v = 300 m/s
• To Find: 1) displacement = y = ?, Velocity of point = v = ?,  Acceleration of point = ?, at x = 1.5 m and t = 0.01 s.

Equation of source is y = 6 sin300πt cm

Comparing with

y = a sin 2πn t cm and y = a sin ω t cm

Amplitude = a = 6 cm = 0.06 m, 2πn = 300 π, n = 150 Hz,  ω = 300 π rad/s

We have v = n λ

∴  λ = v/n  = 300/150 = 2 m

Hence equation of the wave is

∴   y= 0.06 sin 2π(150t – x/2)

∴   y = 0.06 sin 2π(150(0.01) – 1.5/2)

∴  y = 0.06 sin 2π(1.5 – 0.75)

∴  y = 0.06 sin 2π(0.75)

∴  y = 0.06 sin 2π(3/4)

∴  y = 0.06 sin (3π/2)

∴  y = 0.06 (-1) = – 0.06 m

Velocity of particle is given by

Magnitude of the acceleration of particle is given by

f = ω²y = (300 π)² × 0.06 = 5.33 × 104 m/s²

#### Example – 11:

• The equation of simple harmonic oscillations of a source is y = 10 sin 20πt cm. Find the displacement from the position of equilibrium, the velocity and acceleration of point 10 m away from the source 3 seconds after oscillations begin. The velocity of propagation of waves 200 m/s.
• Solution:
• Given: Equation of source is y = 10 sin 20πt cm, v = 200 m/s
• To Find: 1) displacement = y = ?, Velocity of point = v = ?,  Acceleration of point = ?, at x = 10 m and t = 3 s.

Equation of source is y = 10 sin 20πt cm

Comparing with

y = a sin 2πn t cm and y = a sin ω t cm

Amplitude = a = 10 cm = 0.1 m, 2πn = 20 π, n = 10 Hz,  ω = 20 π rad/s

We have v = n λ

∴  λ = v/n  = 200/10 = 20 m

Hence equation of the wave is

∴   y = 0.1  sin 2π(10t – x/20)

∴  y = 0.1  sin 2π(10(3) – 10/20)

∴  y = 0.1 sin 2π(30 – 1/2)

∴  y = 0.1 sin 2π(59/2)

∴  y = 0.1 sin 59π

∴  y = 0.1 (0) = 0 m

Magnitude of the velocity of particle is given by

Magnitude of the acceleration of particle is given by

f = ω²y = (20 π)² × 0 = 0  m/s²

#### Example – 12:

• A transverse simple harmonic progressive wave of amplitude 0.01 m and frequency 500 Hz is traveling along a stretched string with a speed of 200 m/s. Find the displacement of the particle at a distance of 0.7 m from the origin and after 0.01 s.
• Solution:
• Given: Amplitude = a = 0.01 m, frequency = n = 500 Hz, , Velocity of wave = v = 200 m/s
• To Find: 1) displacement = y = ? at x = 0.7 m and t = 0.01 s.

We have v = n λ

∴  λ = v/n  = 200/500 = 0.4 m

Hence equation of the wave is

∴   y = 0.01  sin 2π(500t – x/0.4)

∴   y = 0.01  sin 2π(500(0.01) – 0.7/0.4)

∴   y = 0.01  sin 2π(5 – 7/4)

∴   y = 0.01  sin 2π(13/4)

∴   y = 0.01  sin (13π/2)

∴   y = 0.01  sin (12π + π)/2

∴   y = 0.01  sin (6π + π/2)

∴   y = 0.01  sin (π/2)  = 0.01 × 1 = 0.01 m

 Science > Wave Motion > You are Here

1. Janavie

Can you plz give me solution of electrostatics

• We will post it soon

A wave is represented by the wave equation y=0.5 sin 0.2 π(x-80t ) with all distance in centimetres and time in seconds. Calculate the speed of the wave

• Hemant More

y=0.5 sin 0.2 π(x-80t )
We should convert this equation in form y= a sin 2π(nt – x/λ)
y=0.5 sin π(0.2 x-16t ) ………….. 0.2 taken in bracket
y=0.5 sin 2π(0.2/2 x-16t/2 ) ………….. 2π adjusted out of the bracket and each term is bracket is divided by 2.
y=0.5 sin 2π(0.1 x – 8t) ……………. simplification
y= – 0.5 sin 2π(8t – 0.1x) …………. The terms in bracket interchanged. Note negative sign at start.
y= – 0.5 sin 2π(8t – x/10)
Now comparing with
y= a sin 2π(nt – x/λ)
amplitude = a = 0.5 m, frequency = n = 8 Hz, wavelength = λ = 10 cm
Nov v = nλ = 8 x 10 = 80 cm/s

3. Omar Faruk

X= 0.5Cosπ(100t-Y/0.674), Then,find the velocity of the wave

• Hemant More

X = 0.5 Cos2π(100t-Y/0.674)
X= 0.5Cos2π(50t-Y/1.348)
Comparing with
X= a Cos2π(nt-Y/λ)
n = 50 Hz and λ = 1.348
Now v = nλ = 50 x 1.348 = 67.4 m/s

4. DINESH REDDY

The potential energy of an object is given by U(x) = 8x^2-x^4, Where U is in Joules and x is in meters
a) Determine the force acting on this object?
b) At what position is this object in equilibrium?
c)Which of these equilibrium positions are stable and which are unstable?

• Hemant More

When an object has potential energy, it has a force acting on it (either gravitational, restoring or some other type). The force on the object is the negative derivative of the potential energy function.
F(x) = – (d/dx)F(U) = – (16x -4x^3) = 4x^3 – 16x
When object is at equilibrium F(x) = 0
4x^3 – 16x =0
4x(x^2 – 4) = 0
Hence x = 0 or x = ± 2
Thus at x = 0 and x = ± 2 the object is an equilibrium
Now let us find energy of object at these points
U(0) = 8(0)^2 – (0)^4 = 0
U(±2) = 8(±2)^2 – (±2)^4 = 16
Thus at x = 0 the potential energy is minimum. Hence it is a stable equilibrium.
Similarly, at x = ± 2 the potential energy is maximum. Hence it is an Unstable equilibrium.