[tex]\huge{\bold{\orange{\underline{ Solution }}}}[/tex]
A harmonic wave on a string is described by
[tex]\sf{ Y( x, t) = 0.1 sin(300t + 0.01x + π/3)}[/tex]
Equation for travelling wave :-
[tex]\sf{ Y( x, t) = Asin(ωt + kx + Φ)...eq(1)}[/tex]
Equation for stationary wave :-
[tex]\sf{ Y( x, t) = Acos(ωt - kx )...eq(2)}[/tex]
Given equation for wave :-
[tex]\sf{ Y( x, t) = 0.1 \:sin(300t + 0.01x + π/3)...eq(3)}[/tex]
On comparing eq(1) , (2) and (3)
We can conclude that, Given wave represent travelling wave.
From solution 1 , We can say that,
[tex]\sf{ Y( x, t) = 0.1 \: sin(300t + 0.01x + π/3).}[/tex]
It is travelling from right to left direction
Hence, The direction of its propagation is right to left that is towards +x direction.
Here, We have to find the wave period
We know that,
Wave period = wavelength / velocity
Wave equation :-
[tex]\sf{ Y( x, t) = 0.1 \:sin(300t + 0.01x + π/3).}[/tex]
We know that,
[tex]\sf{v =}{\sf{\dfrac{ ω}{2π}}}{\sf{\: and\:}}{\sf{ λ =}}{\sf{\dfrac{ 2π}{k}}}[/tex]
Subsitute the required values,
[tex]\sf{ wave\: period =}{\sf{\dfrac{ 2π/k}{ω/2π }}}[/tex]
[tex]\sf{ wave \:period = }{\sf{\dfrac{k}{ω}}}[/tex]
[tex]\sf{ wave\: period =}{\sf{\dfrac{ 0.01}{300}}}[/tex]
[tex]\sf{ wave\: period = 0.000033\: s}[/tex]
The wavelength of given wave
[tex]\bold{ λ = }{\bold{\dfrac{2π}{k}}}[/tex]
Subsitute the required values,
[tex]\sf{ λ = }{\sf{\dfrac{2 × 3.14 }{0.01}}}[/tex]
[tex]\sf{ λ = }{\sf{\dfrac{6.28}{0.01}}}[/tex]
[tex]\sf{ λ = 628 \: cm }[/tex]
We have wave equation
[tex]\sf{ Y( x, t) = 0.1 sin(300t + 0.01x + π/3).}[/tex]
Travelling wave equation :-
[tex]\sf{ Y( x, t) = A\:sin(ωt + kx + Φ)...eq(1)}[/tex]
Therefore,
Amplitude of the wave particle
[tex]\sf{ A = 0.1 \: cm}[/tex]
Hence, The amplitude of the particle is 0.1 cm
