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14.Waves and Sound
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A standing wave exists in a string of length $150\ cm$ , which is fixed at both ends with rigid supports . The displacement amplitude of a point at a distance of $10\ cm$ from one of the ends is $5\sqrt 3\ mm$ . The nearest distance between the two points, within the same loop and havin displacment amplitude equal to $5\sqrt 3\ mm$ is $10\ cm$ . Find the maximum displacement amplitude of the particles in the string .... $mm$
A
$20$
B
$15$
C
$10$
D
None of these
Solution
$y = \Lambda \sin kx\,\cos \omega t$
$5 \sqrt{3}=\mathrm{A\,sin\,k} \times 10$
$5 \sqrt{3}=\mathrm{A} \sin \mathrm{k} \times 20$
After solving $\mathrm{k}=\frac{\pi}{30}$
$5\sqrt 3 = {\mathop{\rm Asin}\nolimits} \left( {\frac{{II}}{{30}} \times 10} \right) \to {\rm{A}} = 10\,{\rm{mm}}$
Standard 11
Physics
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