A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, $y(x$, $t )=(0.01 \ m ) \sin \left[\left(62.8 \ m ^{-1}\right) x \right] \cos \left[\left(628 s ^{-1}\right) t \right]$. Assuming $\pi=3.14$, the correct statement$(s)$ is (are) :

$(A)$ The number of nodes is $5$ .

$(B)$ The length of the string is $0.25 \ m$.

$(C)$ The maximum displacement of the midpoint of the string its equilibrium position is $0.01 \ m$.

$(D)$ The fundamental frequency is $100 \ Hz$.

Question

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, $y(x$, $t )=(0.01 \ m ) \sin \left[\left(62.8 \ m ^{-1}\right) x \right] \cos \left[\left(628 s ^{-1}\right) t \right]$. Assuming $\pi=3.14$, the correct statement$(s)$ is (are) :

$(A)$ The number of nodes is $5$ .

$(B)$ The length of the string is $0.25 \ m$.

$(C)$ The maximum displacement of the midpoint of the string its equilibrium position is $0.01 \ m$.

$(D)$ The fundamental frequency is $100 \ Hz$.

IIT 2013, Diffcult

Download our app for free and get started

Solution

$(A)$ There are $5$ complete loops.

Total number of nodes $=6$

$(B)$ $\omega=628 sec ^{-1}$

$k =62.8 m ^{-1}=\frac{2 \pi}{\lambda} \Rightarrow \lambda=\frac{1}{10} $

$v _{ w }=\frac{\omega}{ k }=\frac{628}{62.8}=10 ms ^{-1} $

$L=\frac{5 \lambda}{2}=0.25 $

$(C)$ $2 A =0.01=$ maximum amplitude of antinode

$(D)$ $f=\frac{v}{2 \ell}=\frac{10}{2 \times 0.25}=20 Hz$.

Download our app
and get started for free

Similar Questions

Download our appand get started for free

Similar Questions

Download our app
and get started for free