KVPY SX SB Exam 05-Nov-2017 Question Paper

Equation for a travelling wave is
y=a‌sin(ωt−kx+φ0)
As at t=0 at x=0 and y=0, φ0=0
⇒‌‌φ0=0
So, equation of wave over string is
Here, y‌‌=a‌sin(ωt−kx)
ω‌‌=2πf=2π×500
‌‌=1000π(‌

rad

s

)
Here,
Angular wave number,
k=‌ω∕v=‌1000π∕100=10π(m−1)
So, wave equation is
y=a‌sin(1000π⋅t−10π⋅x)
Given at t=0 at x=0.25m and y=0.02m.
So, ‌‌0.02=a‌sin(−10π×0.25)
=−a‌sin(‌5∕2π)
[∴sin(−θ)=−sin‌θ]
⇒‌‌0.02=−a‌sin(2π+‌π∕2)
⇒‌‌0.02=−a‌sin(‌π∕2) or a=−0.02
Here note that a is amplitude and its positive and negative values are same. When we are getting a negative value this means particle is displaced below mean position.
So, we have
y=−0.02‌sin(1000πt−10πx)
Now, at t=5×10−4s and x=0.2m, value of displacement of particle is
y‌‌=−0.02‌sin(1000π×5×10−4
‌‌=−0.02‌sin(‌

π

2

−2π)
‌‌=0.02‌sin(2π−‌

π

2

)
[∴sin(‌

π

2

−2π)=−sin(2π−‌

π

2

).
‌‌=−0.02m