The number of times y 1 + y 2 = 0 at x = 0 in 1 sec is [2006]

Question

Waves $y_{1} = A \cos (0.5 π x - 100 π t)$ and $y_{2} = A \cos (0.46 π x - 92 π t)$ are travelling along the $x$ -axis. (Here $x$ is in $m$ and $t$ is in second) [2006]

Q. The number of times $y_{1} + y_{2} = 0$ at $x = 0$ in 1 sec is

Smart Achievers · Accepted Answer

(4)

The given equations $y_{1} = A \cos (0.5 π x - 100 π t)$ and $y_{2} = A \cos (0.46 π x - 92 π t)$ represent two progressive waves travelling in the same direction along the $x$ -axis with a slight difference in frequency.

Comparing with the standard equation

$y = A \cos (k x - ω t),$

we get

$ω_{1} = 100 π$ $\Rightarrow 2 π f_{1} = 100 π$ $\Rightarrow f_{1} = 50 Hz$

and

$k_{1} = 0.5 π$ $\Rightarrow \frac{2 π}{λ_{1}} = 0.5 π$ $\Rightarrow λ_{1} = 4 m$

Wave velocity $= λ_{1} f_{1}$ $= 200 m/s$ $[Alternatively, use v = \frac{ω}{k}]$

$ω_{2} = 92 π$ $\Rightarrow 2 π f_{2} = 92 π$ $\Rightarrow f_{2} = 46 Hz$

Therefore, beat frequency $= f_{1} - f_{2}$ $= 4 Hz$ and

$k_{2} = 0.46 π$ $\Rightarrow \frac{2 π}{λ_{2}} = 0.46 π$ $\Rightarrow λ_{2} = \frac{200}{46}$

Wave velocity $= \frac{200}{46} \times 46$ $= 200 m/s$

At $x = 0$ , $y_{1} + y_{2} = (A \cos 100 π t) + (A \cos 92 π t) = 0$

$\Rightarrow \cos 100 π t = - \cos 92 π t = \cos (- 92 π t)$ $= \cos [(2 n + 1) π - 92 π t]$

$\Rightarrow t = \frac{2 n + 1}{192}$

when $t = 0$ , $n = - \frac{1}{2}$ and when $t = 1$ , $n = \frac{191}{2} = 95.5$

$∴$ $In 1 second, y_{1} + y_{2} = 0 at x = 0 for 96 times.$

Solution

Q.
Waves $y_{1} = A \cos (0.5 π x - 100 π t)$ and $y_{2} = A \cos (0.46 π x - 92 π t)$ are travelling along the $x$ -axis. (Here $x$ is in $m$ and $t$ is in second) [2006]

Q. The number of times $y_{1} + y_{2} = 0$ at $x = 0$ in 1 sec is

1 100

2 46

3 192

4 96

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