KVPY SX SB Exam 05-Nov-2017 Question Paper

Question : 31 of 120

Marks: +1, -0

A transverse wave of frequency 500 Hz and speed 100 m/s is traveling in the positive x direction on a long string. At time t = 0 s the displacements at x = 0.0 m and at x = 0.25 m are 0.0 m and 0.02 m, respectively. The displacement at x = 0.2 m at t = 5 × 10

^{–4 }

s is

–0.04 m
–0.02 m
0.04 m
0.02 m

Validate

Solution:

Equation for a travelling wave is

y=a \sin (ω t-k x+φ_{0})

As at

t=0

x=0

and

y=0

φ_{0}=0

⇒ \;\; φ_{0}=0

So, equation of wave over string is
Here,

y \;\; =a \sin (ω t-k x)

ω \;\; =2 π f=2 π × 500

\;\; =1000 π(\;{ {rad}}/{ {s}})

Here,
Angular wave number,

k=\;{ω}\/{v}=\;{1000 π}\/{100}=10 π( {m}^{-1})

So, wave equation is

y=a \sin (1000 π ⋅ t-10 π ⋅ x)

Given at

t=0

x=0.25 {m}

and

y=0.02 {m}

.
So,

\;\; 0.02=a \sin (-10 π × 0.25)

=-a \sin (\;{5}\/{2} π)

[ ∴ \sin (-θ)=-\sin θ]

⇒ \;\; 0.02=-a \sin (2 π+\;{π}\/{2})

⇒ \;\; 0.02=-a \sin (\;{π}\/{2})

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=-0.02 \sin (1000 π t-10 π x)

Now, at

t=5 × 10^{-4} {s}

and

x=0.2 {m}

, value of displacement of particle is

y \;\; =-0.02 \sin (1000 π × 5 × 10^{-4}

\;\; =-0.02 \sin (\;{π}/{2}-2 π)

\;\; =0.02 \sin (2 π-\;{π}/{2})

[ ∴ \sin (\;{π}/{2}-2 π)=-\sin (2 π-\;{π}/{2}) .

\;\; =-0.02 {m}

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