Question

Q] The amplitude of the wave is represented by $\sf y =0.2 \sin \: 4\pi\Bigg[ \dfrac{1}{0.08} - \dfrac{x}{0.8} \Bigg]$ in SI units. Find a) Wavelength b) Frequency c) Amplitude of wave

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Answer 1

Answers :

a) Wavelength = 0.4m

b) Frequency = 25Hz

c) Amplitude = 0.2m

$\rule{300}{2}$

Generalized equation of SHM is :

$\sf y = Asin( \omega t + k) \\ \\ \implies \sf \: y = Asin( \frac{2\pi}{T} t - \dfrac{x}{ \lambda} ) - - - - - - (1)$

Given Equation,

$\sf y =0.2 sin \: 4\pi\Bigg( \dfrac{t}{0.08} - \dfrac{x}{0.8} \Bigg) \\ \\ \implies \sf y =0.2 sin \:2\pi\Bigg( \dfrac{2}{0.08} \times t - \dfrac{2x}{0.8} \Bigg) \\ \\ \implies \sf y =0.2 sin \:2\pi\Bigg( \dfrac{t}{0.04} - \dfrac{x}{0.4} \Bigg) - - - - - - (2)$

Comparing equations (1) and (2),

Wavelength = 0.4m

Amplitude = 0.2m

Time Period = 0.04s

We know that,

$\sf Frequency = \dfrac{1}{Time \ Period}$

Thus,

$\sf \nu = \dfrac{1}{0.04} \\ \\ \longrightarrow \sf \nu = \dfrac{ {10}^{2} }{4} \\ \\ \longrightarrow \boxed{ \boxed{ \sf \nu = 25 {s}^{ - 1} }}$

Answer 2

Solution ⤵

Generalized equation of SHM :

$\sf \: y = Asin ( wt+ k )$

$\sf \: y = Asin ( \frac{2\pi} {Ｔ}t - \frac{x}{λ} )...(1)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \fbox \pink{equation,}$

$\sf \: y = 0.2 \: sin \: 4π ( \frac{t}{0.08} - \frac{x}{0.8} )$

$\sf \: y = 0.2 \: sin \: 2π \: ( \frac{ 2}{0.08} × t - \frac{2x}{0.8} )$

$\sf \: y = 0.2 \: sin \: 2π \: ( \frac{t}{0.04} - \frac{ x}{0.4})...(2)$

$\: \: \: \: \: \: \: \: \: \sf \fbox \red{equation (1) and (2),}$

$\sf\colorbox{pink}{wevelength = 0.4 m}$

$\sf \colorbox{skyblue}{amplitude= 0.2 m}$

$\sf \colorbox{yellow}{timeperiod = 0.04 m}$

So :

$\boxed { \sf \: Frequency = \frac{1}{T.p} }$

Then :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: V = \frac{1}{0.04} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: V = \frac{10²}{4} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \fbox \blue{ \sf \: V = 25ѕ-1}$