The equation for the vibration of a string fixed at both ends vibrating in its third harmonic is given by $y = 2 c m s i n [(0.6 c m^{- 1}) x] c o s [500 π s^{- 1} (t)]$ . The length of the string is -

24.6 c m

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31.4 c m

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20.6 c m

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15.6 c m

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Solution

The correct option is A $15.6 c m$
$y = 2 s i n (0.6 x) c o s (500 π t)$
$ω = 500 π = 2 π f$
$f = 250 H z$
$v = \frac{ω}{k} = \frac{500 π}{0.6}$

Now,
$3 \cdot \frac{v}{2 l} = f$

$o r, \frac{3 \times 500 π}{2 \times l \times 0.6} = 250$

$o r, l = 15.6 c m$

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The equation for the vibration of a string fixed at both ends vibrating in its third harmonic is given by y=2cmsin[(0.6cm−1)x]cos[500πs−1(t)]. The length of the string is -

The correct option is A 15.6cmy=2sin(0.6x)cos(500πt)ω=500π=2πff=250Hzv=ωk=500π0.6Now,3⋅v2l=f or,3×500π2×l×0.6=250or,l=15.6 cm

The equation for the vibration of a string fixed at both ends vibrating in its third harmonic is given by $y = 2 c m s i n [(0.6 c m^{- 1}) x] c o s [500 π s^{- 1} (t)]$ . The length of the string is -

The correct option is A $15.6 c m$
$y = 2 s i n (0.6 x) c o s (500 π t)$
$ω = 500 π = 2 π f$
$f = 250 H z$
$v = \frac{ω}{k} = \frac{500 π}{0.6}$

Now,
$3 \cdot \frac{v}{2 l} = f$

$o r, \frac{3 \times 500 π}{2 \times l \times 0.6} = 250$

$o r, l = 15.6 c m$