Question

The transverse displacement of a string clamped at both ends is given by $y(x,t)=0.06\sin \left(\frac{2\pi x}{3}\right)\cos 60\pi t$, where $x$ and $y$ are in $m$ and $t$ in $s$. The length of the string is $1.5m$ and its mass is $3\times {10}^{-2}kg$. The tension developed in the string is

• A. $81N$
• B. $162N$
• C. $90N$
• D. $180N$

Answer 1

Answer: (B)

$162N$

Given - $m=3\times {10}^{-2}kg,l=1.5m$

Given equation is the equation of a standing wave in a string clamped at both ends , as

$y(x,t)=0.06\sin \left(2\pi x/3\right)\cos 60\pi t$ ....................eq1

The standard equation of standing wave ,

$y(x,t)=2r\sin \left(2\pi x/\lambda \right)\cos \omega t$ ..................eq2

Comparing eq1 and eq2 , we get

$\lambda =3m$

and $\omega =60\pi$

or $2\pi n=60\pi$

or $n=30Hz$

Hence , velocity of wave will be ,

$v=n\lambda$

$v=30\times 3=90m/s$

Now the velocity of standing wave is ,

$v=\sqrt{T/m^{\prime }}$ ..............eq3

where $m^{\prime }$(mass/unit length) $=m/l$

or $m^{\prime }=3\times {10}^{-2}/1.5=2\times {10}^{-2}kg/m$

Hence ,by eq3 ,

$v=\sqrt{T/m^{\prime }}$

or $v^2=T/m^{\prime }$

or $T=v^2{m}^{\prime }$

or $T=(90{)}^2\times 2\times {10}^{-2}=162N$