JEE Advanced 2013 Paper 1

Section: Physics

Question : 12 of 60

Marks: +1, -0

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)×]‌cos[(628s−1)t]

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

[JEE Adv 2013 P1]

The number of nodes is 5.
The length of the string is 0.25 m.
The maximum displacement of the midpoint of the string its equilibrium position is 0.01 m.
The fundamental frequency is 100 Hz.

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