Q.
An open pipe resonates with a tuning fork of frequency 500Hz . It is observed that two successive nodes are formed at distances 16 and 46cm from the open end. The speed of sound in air in the pipe is
Position of first node =16cm 2λ+e=16cm…(i)
Where, e= end correction
Position of second node =46cm 2λ+2λ+e=46cm…(ii)
From Eqs. (i) and (ii), 2λ=30cm λ=60cm =10060m ∴ Speed of sound, V=nλ =500×10060(∵n=500Hz, Given) =300m/s