Question

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The equation for a travelling wave in x - direction is, y = 0.07 sin (80x-3t) where x, y and t are in SI units. Calculate the frequency, velocity and wavelength of the wave.

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Class 11 CBSE

Answer 1

Answer:

Given:

Equation of a travelling wave has been provided :

y = 0.07 sin(80x - 3t)

To find:

Frequency, velocity and wavelength of the wave.

Calculation:

We will first compare the given Equation with a standard Equation of travelling wave :

y = A sin(kx - ωt)

After comparing, we get

• A = 0.07
• k = 80
• ω = 3

1. Velocity of a travelling wave is given as :

v = ω/k

=> v = 3/80

=> v = 0.0375 m/s

2. Wavelength is given as :

k = 2π/λ

=> λ = 2π/k

=> λ = 2π/80

=> λ = π/40 metres.

3. Frequency is given as :

Freq. = ω/2π

=> Freq. = 3/2π Hz.

Answer 2

Question :-

→ Given above

Answer :-

$(1) \: \sf \: v \: = \green{0.038 \: \frac{m}{s}} \: \\ \\ (2) \: \sf \lambda \: = \blue{\frac{ \pi}{40} } \green{or \: } \red{ \frac{22}{280} \: m \: } \\ \\ (3) \: {\sf f = \red{\frac{21}{44}} \: or \: \green{ \frac{3}{2 \pi} } }\sf \: {second}^{ - 1} \: or \: Hz \:$

Explanation :-

Given that ,

→ y = 0.07 sin (80x - 3t)

We know that standard equation of wave ↓

$\leadsto \sf y \red{ = A }\: \green{ \sin(kx - \omega \: t)} \\ \\ \blue{ \sf \: \: after \: comparing \: } \red{\sf \: we \: get \: } \\ \\ \boxed{ \red{\star}} \: \: \sf \: k = 80 \\ \\ \boxed{ \green{\star}} \: \: \sf \omega \: = 3 \\ \\ \boxed{ \purple{ \star}} \sf \: A \: = 0.07$

Now ,

$(1) \sf \: \pink{velocity \: (v) }\\ \\ \implies \sf \red{ v \:} = \green{ \frac{ \omega}{k} } \\ \\ \implies \sf \purple{ v \: = \frac{3}{80} } \\ \\ \implies \boxed{ \sf \: v \: = \green{0.038 \: \frac{m}{s} }} \: \sf (in \: round \: figure) \\ \\ (2) \sf \: \blue{wavelength \: ( \lambda)} \\ \\ \leadsto \sf \red{k = } \green{ \frac{2 \pi}{ \lambda} } \\ \\ \leadsto \sf \lambda \: = \frac{2 \pi}{80} \\ \\ \leadsto \boxed{ \sf \lambda \: = \blue{\frac{ \pi}{40} } \green{or \: } \red{ \frac{22}{280} \: m \: }} \\ \\ (3) \sf \: frequency \: (f) \\ \\ \because \sf \red{ time \:} = \green{ \frac{2 \pi}{ \omega} } \\ \\ \sf \: and \\ \sf \star \: \: frequency \: (f) = \frac{1}{time} \\ \\ \red{ \leadsto \: } \sf \green{ f = } \orange{ \frac{1}{ \red{ \frac{2 \pi}{ \omega} }} } \\ \\ \leadsto \sf \: f = \green{\frac{ \omega}{2 \pi} } \\ \\ \leadsto \sf \red{ f =} \blue{ \frac{3}{2 \pi} } \\ \\ \leadsto \boxed{ \sf f = \red{\frac{21}{44}} \: or \: \green{ \frac{3}{2 \pi} } }\sf \: {second}^{ - 1} \: or \: Hz$

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Hope it helps you .