Two organ pipes open at both ends have same diameter but different lengths. Show that end correction at each end is e=`(n_1l_1-n_2l_2)`/`2(n_2-n_1)` where,the symbols have their usual meaning.

Show that for pipe closed at one end, the end correction is e= (n_1l_1-n_2l_2)/(n_2-n_1) where, the symbols have their usual meaning.

Show that a + r +1=1 where symbols have their usual meaning.

The lengths of two organ pipes open at both ends are L and L + d . If they are sounde together, then the beat frequency will be

State the law of radioactive decay. Hence derive the expression N = N_0(e^-lamda t) .lf where symbols have their usual meanings.

Two organ pipes of the same length open at both ends produces sound of different frequencies , if their radii are different . Why ?

Prove that a pipe open at both end of length of 2L , has same fundamental frequency as another pipe of closed at one end of length L.

Show that σ = \(\frac{ne^2\tau}{m}\). Where symbols have usual meanings.

An organ pipe of length 0.4 m is open at both ends . The speed of sound in air is 340 ms^-1 . The fundamental frequency is

Two vibrating strings of the same material but lengths L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency n_(1) and the other with frequency n_(2) the ratio n_(1)//n_(2) is given by

In a pipe opened at both ends n_(1) and n_(2) be the frequencies corresponding to vibrating lengths L_(1) and L_(2) respectively .The end correction is