MHTCET 14 May 2023 Shift 1 Solved Paper

The given wave equation is:
Y=10sin‌(‌

2πt

30

+α)
At t=0, the displacement Y is given as 5‌cm. So, let's find the phase α using this information:
‌5=10sin‌(α)
‌‌

5

10

=‌

1

2

=sin‌(α)
Since sin‌(30∘)=‌

1

2

, we can say that α=30∘ or α=‌

π

6

radians, considering that sine is positive in the first and second quadrants, and seeing that α represents a phase shift, it usually takes the smallest positive angle that satisfies the equation.
Now, let's calculate the total phase at t=7.5 s :
Total phase =‌

2πt

30

+α
Plug in the values:
Total phase =‌

2π⋅7.5

30

+‌

π

6

Total phase =‌

2π⋅1

4

+‌

π

6

=‌

π

2

+‌

π

6

To add these fractions, we first find a common denominator :
Total phase =‌

3π

6

+‌

π

6

=‌

4π

6

Now we simplify the fraction:
Total phase =‌

2π

3

Therefore, the total phase at t=7.5 s is ‌

2π

3

radians, which corresponds to Option D.