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Distance

A string of l...

Question

A string of length l is fixed at both ends.Its vibrating in its third overtone with maximum amplitude 'a'.The amplitude at distance l/3 from one end is:

\frac{t}{d} = 2

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\frac{t}{d} = 3

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\frac{t}{d} = 6

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(\sqrt{3 / 2}) a

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Solution

The correct option is B $(\sqrt{3 / 2}) a$
Clearly its vibrating in third overtone there we will get nodes at $x = 0, \frac{L}{4}, \frac{2 L}{4}, \frac{3 L}{4}$
The equation of motion is $y = a sin (\frac{4 x}{L}) cos x t$
put $x = \frac{1}{3}$
Amplitude $= a sin (\frac{4 L}{3 L})$
$= a sin (\frac{4}{3})$
$= a (\frac{v^{3}}{2})$
so, $(\frac{\sqrt{3}}{2}) a,$ D option is correct.
Hence, the answer is $(\frac{\sqrt{3}}{2}) a .$

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A string of length l is fixed at both ends.Its vibrating in its third overtone with maximum amplitude 'a'.The amplitude at distance l/3 from one end is:

The correct option is B (√3/2)aClearly its vibrating in third overtone there we will get nodes at x=0,L4,2L4,3L4The equation of motion is y=asin(4xL)cosxtput x=13Amplitude =asin(4L3L) =asin(43) =a(v32)so, (√32)a, D option is correct. Hence, the answer is (√32)a.