Wave Motion 2 Question 30

30. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x,t)=(0.01m)[sin(62.8m1)x]cos[(628s1)t].

Assuming π=3.14, the correct statement(s) is (are)

(2012)

(a) the number of nodes is 5

(b) the length of the string is 0.25m

(c) the maximum displacement of the mid-point of the string from its equilibrium position is 0.01m

(d) the fundamental frequency is 100Hz

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Answer:

Correct Answer: 30. (b, c)

Solution:

Number of nodes =6

From the given equation, we can see that

k=2πλ=62.8m1λ=2π62.8m=0.1ml=5λ2=0.25m

The mid-point of the string is P, an antinode maximum displacement =0.01m

ω=2πf=628s1f=6282π=100Hz

But this is fifth harmonic frequency.

Fundamental frequency f0=f5=20Hz



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