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#### 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 × 10^{4} 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 > Physics > Wave Motion > You are Here |

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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

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

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

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

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?

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.