Question

A horizontal stretched string, fixed at two ends, is vibrating in its $5^{th}$ harmonic according to the equation, $y(x,t)=\left(0.10m\right)\sin \left[\right(62.8m^{-1}\left)x\right]\cos \left[\right(628s^{-1}\left)t\right]$. Assuming $\pi =3.14$, the correct statement(s) is/are :

• A. The number of nodes is $5$
• B. The length of the string is $0.25m$
• C. The maximum displacement of the midpoint of the string, from its equilibrium position is $0.01m$
• D. The fundamental frequency is $100Hz$

Answer 1

Answer: (A), (B)

The length of the string is $0.25m$
The number of nodes is $5$

According to question we have

No of nodes $=5$

equation of wave : $y=\left(0.1\right)sin\left(62.8x\right)cos\left(628t\right)$

From the equation wave equation.

$K=62.8$

$\Rightarrow \frac{2\pi }{\lambda }=62.8$

$\lambda =\frac{2\times 3.14}{62.8}=0.1m$

Now,

For $5^{th}$ Harmonic motion

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

$\Rightarrow L=\frac{5\times 0.1}{2}=0.25m$ (Length of the string)

Amplitude, i.e. maximum displacement $=0.1m$

Fundamental frequency, $F=\frac{1}{2L}\sqrt{\frac{T}{\mu }}OR\frac{W/K}{2L}$

$=\frac{628/62.8}{2\times 0.25}=20Hz$

$\therefore$ Option (A) & (B) are correct.