Question

A string of length $1.5\text{m}$ with both its ends clamped is vibrating in fundamental mode. Amplitude at the centre of string is $4\text{mm}$. What is the minimum distance between two points having amplitude $2\text{mm}$?

• A. $1\text{m}$
• B. $1.25\text{m}$
• C. $0.75\text{m}$
• D. $0.50\text{m}$

Answer 1

The correct option is A $1\text{m}$

As the string is vibrating in fundamental mode, therefore, the distance between the two nodes will be equal to the length of the string.
$\Rightarrow \frac{\lambda }{2}=1.5$
$\Rightarrow \lambda =3\text{m}$
We know the equation of standing wave for string having both end fixed,
$y=2A\sin \left(kx\right)\cos \left(\omega t\right)$
At a particular instant, the amplitude of a particle at a distance $x$ from one of the ends,
$y=2A\sin \left(kx\right)$
$\Rightarrow 2=4\sin \left(kx\right)$ [given]
$\Rightarrow \sin \left(kx\right)=\frac{1}{2}$
$\Rightarrow kx=\frac{\pi }{6}$ or $kx=\frac{5\pi }{6}$
$\Rightarrow \frac{2\pi x}{\lambda }=\frac{\pi }{6}$ or $\frac{2\pi x}{\lambda }=\frac{5\pi }{6}$
$\Rightarrow x=\frac{\lambda }{12}$ or $x=\frac{5\lambda }{12}$
$\Rightarrow x=\frac{3}{12}$ or $x=\frac{5\times 3}{12}$
$\Rightarrow x=0.25\text{m}$ or $x=1.25\text{m}$
Therefore,
$\Delta x=1.25-0.25=1\text{m}$