# Wave and motion

### Numerical

Part I

1.

Equation of harmonic wave y = 4.0sin$\left( {{\rm{\pi }}\left( {2.0{\rm{t}} - 0.01{\rm{x}}} \right)} \right)$ne

Comparing this equation with y = Asin(wt – kx) we get,

(i) Amplitude = 4cm

(ii) Wavelength = k = $\frac{{2{\rm{\pi }}}}{\lambda }$

Or, 0.01n = $\frac{{2{\rm{n}}}}{\lambda }$

So, λ = 200cm = 2m

(iii) Initial phase θo = 0 [at t = 0]

(iv) For speed of wave,

W = 2.0π

Or, 2πf = 2.0π

Or, $\frac{{\rm{v}}}{\lambda }$ = 1

So, v = 2m/s.

(v) w = 2.0π

2πf = 2.0π

So, f = 1Hz.

2.

Soln:

y = 0.25 * 10-3 sin(500t – 0.025x)

Comparing this equation with y = A sin(wt – kx)

We get, amplitude = 0.25 * 10-3cm

or, $\frac{{2{\rm{\pi }}}}{{\rm{T}}}$ = 500

So, T = $\frac{{250}}{{\rm{\pi }}}$ sec.

So, k = $\frac{{2{\rm{\pi }}}}{\lambda }$

So, λ = 251.32cm

Particle velocity (v) = $\frac{{{\rm{dy}}}}{{{\rm{dT}}}}$ = 0.25 * 10-3 * 50Cos (500 + - 0.025x)

So, V = 0.125Cos(500 + - 0.025x)

So, amplitude of particle velocity (A) = 0.125 cm/s.

Acceleration (a) = $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = 0.125 * 500sin(500 + - 0.0025x)

= 62.5sin(500 + - 0.0025x)

So, amplitude of acceleration (A) = 62.5 cm/s2.

3.

Soln:

y = 10-6sin(100t + 20x + $\frac{{\rm{\pi }}}{{\rm{a}}}$)

Comparing with y = A sin(wt + kx + θ)

We get,

K = 20,

Or, $\frac{{2{\rm{\pi }}}}{\lambda }$ = 20

So, w = 100

Or, 2πf = 100

Or, 2π$\frac{{\rm{v}}}{\lambda }$ = 100

So, v = 5m/s.

4.

Soln:

Equation of travelling wave,

Y = 10sinπ(0.01x – 2t)

Comparing it with y = Asint(wt – kx)

Y = Asin(kx – wy) [sin(-θ) = sinθ]

So, Amplitude (A) = 10cm

k  = 0.01π

or, $\frac{{2{\rm{\pi }}}}{\lambda } = 0.01{\rm{*n}}$

Wavelength (λ) = 200cm = 2m

Again,

w = 2

or, 2πf = 2π

So, f = 1Hz.

Velocity = Frequency * Wavelength

= 1 * 2 = 2m/s.

Part II

5.

Soln:

Velocity of Ripple(v) = 0.20m/s.

λ = 15 * 10-3 m.

Amplitude (A) = 5.0 * 10-3m

Equation of S.H. wave,

y = Asin(wt – kx)

= Asin$\left( {\frac{{2{\rm{nv}}}}{\lambda }{\rm{t}} - \frac{{2{\rm{nx}}}}{\lambda }} \right)$

= A.sin$\frac{{2{\rm{n}}}}{\lambda }$(vt – x)

= Asin$\frac{{2{\rm{\pi }}}}{\lambda }$(vt – kx)

Velocity of particle (v) = $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$

= A * $\frac{{2{\rm{nv}}}}{\lambda }{\rm{cos}}\left( {{\rm{vt}} - {\rm{kx}}} \right)$

At maximum velocity, vt – kx = 0

Vmax = A.$\frac{{2{\rm{nv}}}}{\lambda }$

= $\frac{{5{\rm{*}}{{10}^{ - 3}}{\rm{*}}2{\rm{*n*}}0.20}}{{15{\rm{*}}{{10}^{ - 3}}}}$

= 0.42m/s.

6.

Soln:

Velocity of sound in air = 340m/s.

When f = 256Hz.

Wavelenght (λ) = $\frac{{{\rm{velocity}}}}{{{\rm{frequency}}}}{\rm{\: }}$

= $\frac{{340}}{{256}}$ = 1.32m

When λ = 0.85m

Frequency = $\frac{{{\rm{velocity}}\left( {\rm{v}} \right)}}{{{\rm{wavelength}}\left( \lambda \right)}}$

= $\frac{{340}}{{0.85}}$ = 400Hz.

7.

Equation of plane progressive wave y = asin(wt – kx)

Here, wavelength = $\frac{{{\rm{velocity}}}}{{{\rm{frequency}}}}$ = $\frac{{30}}{{250}}$ = 0.12m

Phase difference between two points when x = 0.1m apart.

θ = $\frac{{2{\rm{n}}}}{\lambda }$x = $\frac{{2{\rm{n}}}}{{0.12}}$ * 0.1 = $\frac{{5{\rm{n}}}}{3}$

So, Equation of plane progressive wave,

y = 0.03sin(2n * 250t – $\frac{{2{\rm{n}}}}{{0.12}}$ * x)

= 0.03sin2n(250t – $\frac{{25}}{3}$x)

So, Distance between two nodes = $\frac{\lambda }{2}$ = $\frac{{0.12}}{2}$ = 6cm.

8.

Soln:

Equation of stationary waves,

y = 2acos$\frac{{2{\rm{n}}}}{\lambda }$ * sin$\frac{{2{\rm{n}}}}{\lambda }$t

Incomplete

9.

10.

y = 0.1sin$\left( {200{\rm{nt}} - \frac{{200}}{{17}}{\rm{nx}}} \right)$

Comparing it with y = asin(wt – kx)

W = 200n

2πF = 200π

So, Frequency (f) = 100Hz.

Also, $\frac{{200{\rm{\pi }}}}{{17}} = {\rm{k}}$

Or, $\frac{{2{\rm{\pi }}}}{\lambda } = \frac{{200{\rm{\pi }}}}{{17}}$

So, λ = $\frac{{17}}{{100}}$ = 0.17cm = 1.7m

For speed of wave,

Or, $\frac{{2{\rm{nv}}}}{\lambda }$ = 200n

Or, v = λ * 100

So, v = 1.7 * 100 = 170m/s.

Phase difference θ between two points i.e. at x1 = 1.10m and x2 = 0.25m is given by θ = $\frac{{2{\rm{n}}}}{\lambda }$(x1 – x2)

Or, $\frac{{2{\rm{n}}}}{{1.7}}\left( {1.10 - 0.25} \right)$ = π

When amplitude is doubled A = 2 * 0.1 = 0.2

Frequency is doubled F = 2 * 100Hz = 200Hz.

Wave travelling in opposite direction is given by:

y  = Asin(wt + kx)

= 0.2sin(2π * 200t + $\frac{{200}}{{17}}$nx)

11.

Given:

y = Asin(wt – kx)

a. y = Asin(wt - $\frac{{2{\rm{n}}}}{\lambda }{\rm{x}}$)

= Asin(wt – wx)

= A sinw(t – $\frac{{\rm{x}}}{{\rm{v}}}$) [where $\frac{{2{\rm{n}}}}{\lambda } = \frac{{2{\rm{nF}}}}{{\rm{v}}}$]

So, 2πF = w

Here, x is the time taken by the wave to travel from origin to  x in + vediraction and t – $\frac{{\rm{x}}}{{\rm{v}}}$ is the motion of wave at earlier time which is equal to w motion of wave at point x at tome ‘t’.

For speed of the wave,

Or, k = $\frac{{2{\rm{n}}}}{\lambda }$

Or, k = $\frac{{2{\rm{\pi }}}}{{\rm{v}}}{\rm{*f}} = \frac{{\rm{w}}}{{\rm{v}}}$

So, v = $\frac{{6.6{\rm{*}}{{10}^3}}}{{20}}$ = 330m/s.

From the given information we get,

Y = 1.0 * 10-7sin(6.6 * 103t – 20x)

Spped of particle u = $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$

= 1.0 * 10-7 * 6.6 * 103cos(6.6 * 103t – 20x)

= 6.6 * 10-4cos(6.6 * 103t – 20x)

For maximum speed, we get, wt – kx = 0

umax = 6.6 * 10-4 m/s.

Part-iii

12.

Speed of sound (v) = 344m/s

Frequency (F) = 262Hz

Wavelength = $\frac{{{\rm{velocity}}}}{{{\rm{Frequency}}}}$ =  $\frac{{344}}{{262}}$

= 1.31m

13.

F = 2Hz

Amplitude (a) = 7.5cm = 0.075m.

Wave speed (v) = 12m/s

So, Angular frequency (w) = 2πf

= 2π * 2 = 4π = 12.57 rad/s/

Wavelength = $\frac{{\rm{v}}}{{\rm{F}}}$ = $\frac{{12}}{2}$ = 6m

Wave number = $\frac{{2{\rm{n}}}}{\lambda }$ = $\frac{{2{\rm{n}}}}{6}$ = $\frac{{\rm{n}}}{3}$ = 1.047 rad/m.

Equation of wave,

y(x1t) = 0.075sin(wt – kx)

= 0.075(sinw(t – $\frac{{\rm{x}}}{{12}}$))

= 0.075sin12.75(t – $\frac{{\rm{x}}}{{12}}$)

14.

Speed of wave (v) = 143m/s.

Frequency (f) = 440Hz.

Amplitude(a) = 0.75mm = 0.75 * 10-3m.

Equation of standing waves,

(i) y = 2acoskx.sinwt

= 2 * 0.75 * 10-3 * cos$\frac{{2{\rm{n}}}}{\lambda }$x . sin2ft

= 1.5 * 10-3cos 2π * $\frac{{\rm{x}}}{{0.325}}$ sin2π * 440t

= 1.5 * 10-3cos 19.33x sin2760t

(ii) Amplutde = 1.5 * 10-3m

Vmax = $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ at x = 0 and t = 0.

= 1.5 * 10-3 * 2760 = 4.15m/s.

15.

Wavelength of wave = distance between crests = 6m

Time taken by boat to complete one wave = 2.5 * 25 = 5 sec.

Wave velocity = $\frac{{\rm{w}}}{{\rm{k}}}$ = $\frac{{2{\rm{\pi }}}}{{{\rm{T*}}\frac{{2{\rm{\pi }}}}{\lambda }}}$ = $\frac{\lambda }{{\rm{T}}}$.

= $\frac{6}{5}$ = 1.2m/s.

16.

Given transverse wave,

y(x1t) = 0.65sin2π($\frac{{\rm{t}}}{{0.036}} - \frac{{\rm{x}}}{{28}}$)

y(x1t) = 0.65sin$\left( {\frac{{2{\rm{\pi t}}}}{{0.036}} - \frac{{2{\rm{\pi x}}}}{{28}}} \right)$

comparing it with y = asin(wt – kx)

Amplitude = 0.65cm = 6.5 * 10-3m

k = $\frac{{2{\rm{\pi }}}}{\lambda }$

or, $\frac{{2{\rm{\pi }}}}{{28}}$ = $\frac{{2{\rm{\pi }}}}{\lambda }$

Wavelegnth (λ) = 28cm = 28 * 10-2m

Also,

w = $\frac{{2{\rm{\pi }}}}{{0.036}}$

So, Frequency (F) = 27.78Hz.

Wave number (k) = $\frac{{2{\rm{\pi }}}}{\lambda }$ = $\frac{{2{\rm{\pi }}}}{8}$

Speed of propagation (v) = $\frac{{\rm{w}}}{{\rm{k}}}$

= $\frac{{\frac{{2{\rm{\pi }}}}{{0.036}}}}{{\frac{{2{\rm{\pi }}}}{{28}}}}$

= 7.77m/s.

17.

Soln:

Wave speed (v) = 8m/s.

Amplitude (a) = 0.07 m/s.

Wavelength (λ) = 0.32m

(i).

Frequency = $\frac{{{\rm{wave\: speed}}}}{{{\rm{wavelength}}}}$

= $\frac{8}{{0.32}}$

= 25Hz.

Period = $\frac{1}{{\rm{F}}}$ = $\frac{1}{{25}}$ = 0.045

Wave number (k) = $\frac{{2{\rm{\pi }}}}{\lambda }$ = $\frac{{2{\rm{\pi }}}}{{0.32}}$

(ii)

y = 0.007 sin(wt – kx)

= 0.07sin2π $\left( {\frac{{\rm{t}}}{{\rm{F}}} - 19.63{\rm{x}}} \right)$

= 0.07 sin2π $\left( {\frac{{\rm{t}}}{{0.04}} - 19.63{\rm{x}}} \right)$

(iii)

At t = 0.15sec and x = 0.36

y = 0.07sin2π$\left( {\frac{{0.15}}{{0.04}} - 19.63{\rm{*}}0.36} \right)$

= -0.0495 m.

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