Question
Question: An open tube is in resonance with string ( frequency of vibration of tube is nₒ) . If tube is dipped...
An open tube is in resonance with string ( frequency of vibration of tube is nₒ) . If tube is dipped in water so that 75% of the length of tube is inside water, then the ratio of frequency of tube to the string now will be,
& \text{A}\text{. }\dfrac{l}{2} \\\ & \text{B}\text{. 2} \\\ & \text{C}\text{. }\dfrac{2}{3} \\\ & \text{D}\text{. }\dfrac{3}{2} \\\ \end{aligned}$$Solution
Fundamental frequency for open tube concept is used to solve the above problem.
Formula used: Fundamental frequency for open tube η0=2lv where, v=velocity of sound and l = length of the tube.
And, frequency for the tube which is closed on one side and open on the other side = η0=4lv
Complete step by step solution:
The fundamental frequency for open tube is given by,
η0=2lv...............(i), when we are assuming that tube is open.
When the tube is dipped inside water, it becomes closed on one side and open on the other side, therefore, length of the tube
available as closed tube is ,
l1=(1−10075)x lwhere, l₁is the length for the part of the tube closed on one side and open on the other side.
Therefore, l1=(10025)x l=4l
Now, we know that frequency for the tube which is closed on one side and open on the other side is given by,
η=4l1v=4x(4l)v=lv................(ii)
Let us compare equations (i) and (ii) we get,