JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 9)
A string is clamped at both the ends and it is
vibrating in its 4th harmonic. The equation of the
stationary wave is Y = 0.3 sin(0.157x) cos(200pt).
The length of the string is : (All quantities are
in SI units.)
60 m
20 m
80 m
40 m
Explanation
4th harmonic
$$4{\lambda \over 2} = l;2\lambda = l$$
From equation $${{2\pi } \over \lambda } = 0.157$$
$$\lambda $$ = 40 ; $$l$$ = 2$$\lambda $$ = 80 m
$$4{\lambda \over 2} = l;2\lambda = l$$
From equation $${{2\pi } \over \lambda } = 0.157$$
$$\lambda $$ = 40 ; $$l$$ = 2$$\lambda $$ = 80 m
Comments (0)
