Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A string is clamped at both the ends and it is vibrating in its $4^{th}$ harmonic. The equation of the stationary wave is $Y=0.3\: sin\left ( 0.157x \right )cos\left ( 200\pi t \right ).$ The length of the string is : (All quantities are in $SI$ units.)

Option 1)
$20\: m$

Option 2)
$80\: m$

Option 3)
$40\: m$

Option 4)
$60\: m$

Answers (1)

best_answer

$y=0.3\; sin\left ( 0.157x \right )cos\left ( 200\pi t \right )$

$y=2A sin \left ( kx \right )cos\left ( wt \right )$

$k=\frac{2\pi}{\lambda }$

$\lambda =\frac{2\pi }{K}$

$4\left ( \frac{\lambda }{2} \right )=l$

$l\Rightarrow 2\times \lambda$

$l\Rightarrow 2\times \frac{2\pi }{K}$

$=2\times \frac{2\times 3.14}{\cdot 157}$

$=80\; m$

Option 1)

$20\: m$

Option 2)

$80\: m$

Option 3)

$40\: m$

Option 4)

$60\: m$

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs

anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question