JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Waves & Sound

The vibrator drives standing waves on the fixed string, and the number of loops follows from $f=\frac{n}{2\ell }\sqrt{\frac{T}{\mu }}$. With $\mu =\frac{0.005}{1}=0.005$ kg/m, $\sqrt{T/\mu }=\sqrt{8.0/0.005}=\sqrt{1600}=40$ m/s, so $100=\frac{n}{2}\times 40=20n$ gives $n=5$ loops along the one-metre string. Successive nodes are half a wavelength apart, which is one loop length, $\frac{1}{5}$ m, that is $20$ cm. The frequency fixes the wave speed and so how many half-wavelengths the string can hold.