Question

plzz help me in the problem !!!​

Answer 1

⍟ Plane Progressive Waves ⍟

• A plane progressive wave is one that moves in space, but lies in a single plane all the time. It has a constant amplitude and a single direction of propagation.

• The Particle Displacement y for a plane progressive wave can be represented by the following equation:

$y=A\sin(\omega t-kx)$

where

$A = \textsf{Amplitude} \\ \\ \omega = \textsf{Angular Frequency} = 2\pi f \\\\ k = \textsf{Angular Wave Number} = \frac{2\pi}{\lambda}$

Taking the $2\pi$ factor common, we have:

$\begin{aligned}\rightarrow y &= A\sin(\omega t-kx) \\\\\\ &= A\sin \left(2\pi ft - \dfrac{2\pi}{\lambda}x\right) \\\\\\ \therefore y &= A\sin 2\pi \left(ft-\frac{1}{\lambda}x\right)\end{aligned}$

• Now, Let's take a look at our given equation:

$y=10\sin 2\pi (t-0.005x) \\\\\\ \textsf{Comparing with standard form } {\sf y=A\sin 2\pi\left(ft-\dfrac{1}{\lambda}x\right)} \\\\\\\Large \rightarrow \boxed{A = 10 \: cm} \\\\\\ \rightarrow \boxed{f = 1\ s^{-1} = 1\ Hz} \\\\\\ \rightarrow \dfrac{1}{\lambda}=0.005\ cm^{-1} \\\\\\ \implies \lambda = \dfrac{1}{0.005} = 200\ cm = 2\ m \\\\\\ \implies \boxed{\lambda=2\ m}$

• Also, Wave Velocity, say $v$, is given by:

• $v = \lambda f \\\\\\ \implies v = 2\times 1\ m/s \\\\\\ \implies \Large \boxed{v=2\ m/s}$

• Thus, we found everything we needed. Here's the summary of final results:

$\large\underline{\texttt{SUMMARY OF RESULTS}}\\\boxed{\begin{minipage}{20em}\tt Amplitude = A = 10 cm \\\\\\ Frequency = f = 1 Hz \\\\\\ Wavelength = \lambda = 200 cm = 2 m\\\\\\ Wave Velocity = v = 200 cm/s = 2 m/s \end{minipage}}$

Answer 2

Plane Progressive Waves ⍟

A plane progressive wave is one that moves in space, but lies in a single plane all the time. It has a constant amplitude and a single direction of propagation.

The Particle Displacement y for a plane progressive wave can be represented by the following equation:

y=A\sin(\omega t-kx)y=Asin(ωt−kx)

where

\begin{gathered}A = \textsf{Amplitude} \\ \\ \omega = \textsf{Angular Frequency} = 2\pi f \\\\ k = \textsf{Angular Wave Number} = \frac{2\pi}{\lambda}\end{gathered}

A=Amplitude

ω=Angular Frequency=2πf

k=Angular Wave Number=

λ

2π

Taking the 2\pi2π factor common, we have:

\begin{gathered}\begin{aligned}\rightarrow y &= A\sin(\omega t-kx) \\\\\\ &= A\sin \left(2\pi ft - \dfrac{2\pi}{\lambda}x\right) \\\\\\ \therefore y &= A\sin 2\pi \left(ft-\frac{1}{\lambda}x\right)\end{aligned}\end{gathered}

→y

∴y

=Asin(ωt−kx)

=Asin(2πft−

λ

2π

x)

=Asin2π(ft−

λ

1

x)

Now, Let's take a look at our given equation:

\begin{gathered}y=10\sin 2\pi (t-0.005x) \\\\\\ \textsf{Comparing with standard form } {\sf y=A\sin 2\pi\left(ft-\dfrac{1}{\lambda}x\right)} \\\\\\\Large \rightarrow \boxed{A = 10 \: cm} \\\\\\ \rightarrow \boxed{f = 1\ s^{-1} = 1\ Hz} \\\\\\ \rightarrow \dfrac{1}{\lambda}=0.005\ cm^{-1} \\\\\\ \implies \lambda = \dfrac{1}{0.005} = 200\ cm = 2\ m \\\\\\ \implies \boxed{\lambda=2\ m}\end{gathered}

y=10sin2π(t−0.005x)

Comparing with standard form y=Asin2π(ft−

λ

1

x)

→

A=10cm

→

f=1 s

−1

=1 Hz

→

λ

1

=0.005 cm

−1

⟹λ=

0.005

1

=200 cm=2 m

⟹

λ=2 m

Also, Wave Velocity, say vv , is given by:

\begin{gathered}v = \lambda f \\\\\\ \implies v = 2\times 1\ m/s \\\\\\ \implies \Large \boxed{v=2\ m/s}\end{gathered}

v=λf

⟹v=2×1 m/s

⟹

v=2 m/s