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(−‌). This can be rewritten as
y=3sin‌(−2π‌), which matches the form
y=Asin‌(ωt−kx) where
ωt is zero (i.e.,
t=0),A is 3 and
k is
−‌.
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=‌.
4. The speed of the wave
v is given by
v=‌. From the problem, we know that
v=4m∕ s. By substituting these values, we can solve for
ω :
4=‌ which gives
ω=‌.
5. So, the wave equation at any time
t can be written as
y=3sin‌[‌t−‌x].
6. Now, we want to find the wave equation at
t=4 s. Substituting
t=4 into the equation gives
y=3sin‌[‌×4−‌x].
7. Simplifying the argument of the sine function, we get
y=3sin‌2π(‌).
8. Rearranging the argument of the sine function again, we get
y=3sin‌2π(−‌).