Wave Motion 2 Question 3

3. A string is clamped at both the ends and it is vibrating in its 4 th harmonic. The equation of the stationary wave is Y=0.3sin(0.157x)cos(200πt). The length of the string is (All quantities are in SI units)

(Main 2019, 9 April I)

(a) 60m

(b) 40m

(c) 80m

(d) 20m

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Answer:

Correct Answer: 3. (c)

Solution:

  1. Given equation of stationary wave is

Y=0.3sin(0.157x)cos(200πt)

Comparing it with general equation of stationary wave, i.e. Y=asinkxcosωt, we get

k=(2πλ)=0.157λ=2π0.157=4π2(12π0.157). and ω=200π=2πTT=120s

As the possible wavelength associated with nth harmonic of a vibrating string, i.e. fixed at both ends is given as

λ=2ln or l=n(λ2)

Now, according to question, string is fixed from both ends and oscillates in 4th harmonic, so

4(λ2)=l2λ=l

or l=2×4π2=8π2 [using Eq. (i)]

Now, π210l80m



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