A string is stretched between two rigid supports separated by $$75 \mathrm{~cm}$$. There are no resonant frequencies between $$420 \mathrm{~Hz}$$ and $$315 \mathrm{~Hz}$$. The lowest resonant frequency for the string is
A progressive wave is given by, $$\mathrm{Y}=12 \sin (5 \mathrm{t}-4 \mathrm{x})$$. On this wave, how far away are the two points having a phase difference of $$90^{\circ}$$ ?
The equation of the wave is $$\mathrm{Y}=10 \sin \left(\frac{2 \pi \mathrm{t}}{30}+\alpha\right)$$ If the displacement is $$5 \mathrm{~cm}$$ at $$\mathrm{t}=0$$ then the total phase at $$\mathrm{t}=7.5 \mathrm{~s}$$ will be $$\left(\sin 30^{\circ}=0.5\right)$$
If '$$l$$' is the length of the open pipe, '$$r$$' is the internal radius of the pipe and '$V$ ' is the velocity of sound in air then fundamental frequency of open pipe is