Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Find the number of times ${y_1} + {y_2} = 0$ at $x = 0$ in 1sec.A.100B.46C.192D.96

Last updated date: 13th Jun 2024
Total views: 393k
Views today: 7.93k
Verified
393k+ views
Hint: We know that the phase of a wave specifies the location of a point within a wave cycle of a repetitive waveform. Here, we use the expression of displacement y of a cosine wave having direction of propagation, phase. In this relation, we put the observed values from the given figure and get the required result.
Formula used:
$y = A\cos \left[ {kx - \omega t} \right]$

Here, we have two wave equations, where ${y_1},{y_2}$ are representing the wave displacement, A is the amplitude of the wave, k is the wave vector, x is position, t is the time.
$y = A\cos \left[ {kx - \omega t} \right]$
Now we will use the above relation and put the given values, that are x=0 at t=1 sec.
\eqalign{ & {y_1} = A\cos \left[ {0.5\pi \times 0 + 100\pi } \right] \cr & \Rightarrow {y_1} = A\cos 100\pi \cr}
For second wave we have:
\eqalign{& {y_2} = A\cos \left[ {0.46\pi \times 0 - 92\pi } \right] \cr & \Rightarrow {y_1} = A\cos ( - 92\pi ) \cr & \Rightarrow {y_1} = A\cos 92\pi \cr}
We have the relation:
${y_1} + {y_2} = 0$
By substituting the values in above equation, we get:
$\Rightarrow A\left[ {cos100\pi + \cos 92\pi } \right] = 0$
$\therefore A2\sin 96\pi sin4\pi = 0$

Therefore, we get the required result and the correct option is A) i.e., the number of times in given condition is 100.