Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2013-Paper-1 A horizontal stretched string fixed at two e…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2013-Paper-1 Single Correct MCQ
Published on: August 13, 2026

A horizontal stretched string fixed at two ends, is vibrating in its fifth harmonic according to the equation y(x, t) = 0.01m sin [(62.8m -1 )x] cos[(628s -1 )t]. Assuming = 3.14, the correct statement(s) is (are)

A
The number of nodes is 5.
B
the length of the string is 0.25 m.
C
The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01m.
D
The fundamental frequency is 100 Hz.

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Text Solution

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The correct answer is:
CHECK THE SOLUTION.

(B, C)

y = 0.01 m sin (20 x) cos 200 t

no. of nodes is 6

length of the spring = 0.5 x = 0.25

Mid point is the antinode

Frequency at this mode is f =

Fundamental frequency = 20Hz.

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