WBJEE 2023 Physics & Chemistry Solved Paper

1. The general equation of a wave propagating in one dimension can be expressed as y=Asin‌(ωt−kx+φ), where A is the amplitude, ω is the angular frequency, k is the wave number, x is the position, t is the time, and φ is the phase constant.
2. The given wave equation is y=3sin‌2π(−‌

x

3

). This can be rewritten as y=3sin‌(−2π‌

x

3

), which matches the form y=Asin‌(ωt−kx) where ωt is zero (i.e., t=0),A is 3 and k is −‌

2π

3

.
3. The negative sign of k indicates that the wave is moving in the negative x-direction, but the problem states that the wave is moving in the positive x-direction. Hence, we should take k as positive, i.e., k=‌

2π

3

.
4. The speed of the wave v is given by v=‌

ω

k

. From the problem, we know that v=4m∕ s. By substituting these values, we can solve for ω : 4=‌

ω

2π∕3

which gives ω=‌

8π

3

.
5. So, the wave equation at any time t can be written as y=3sin‌[‌

8π

3

t−‌

2π

3

x].
6. Now, we want to find the wave equation at t=4 s. Substituting t=4 into the equation gives y=3sin‌[‌

8π

3

×4−‌

2π

3

x].
7. Simplifying the argument of the sine function, we get y=3sin‌2π(‌

−x+16

3

).
8. Rearranging the argument of the sine function again, we get y=3sin‌2π(−‌

x−16

3

).