SATHEE: Wave Motion 1 Question 13

Wave Motion 1 Question 13

13. A wave equation which gives the displacement along the $y$ -direction is given by : $y = 10^{- 4} \sin (60 t + 2 x)$ where, $x$ and $y$ are in metre and $t$ is time in second. This represents a wave

(1981, 3M)

(a) travelling with a velocity of $30 m / s$ in the negative $x$ -direction

(b) of wavelength $(π) m$

(d) of amplitude $10^{- 4} m$

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Answer:

Correct Answer: 13. $(a, b, c, d)$

Solution:

$y = 10^{- 4} \sin (60 t + 2 x)$

$A = 10^{- 4} m, ω = 60 r a d / s, k = 2 m^{- 1}$

Speed of wave, $v = \frac{ω}{k} = 30 m / s$

Frequency, $f = \frac{ω}{2 π} = \frac{30}{π} H z$ .

Wavelength $λ = \frac{2 π}{k} = π m$

Further, $60 t$ and $2 x$ are of same sign. Therefore, the wave should travel in negative $x$ -direction.

$∴$ All the options are correct.

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Wave Motion 1 Question 13

13. A wave equation which gives the displacement along the y-direction is given by : y=10−4sin⁡(60t+2x) where, x and y are in metre and t is time in second. This represents a wave

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13. A wave equation which gives the displacement along the $y$ -direction is given by : $y = 10^{- 4} \sin (60 t + 2 x)$ where, $x$ and $y$ are in metre and $t$ is time in second. This represents a wave