Question

An air column closed at one end and open at the other end resonates with a tuning fork when the length of the column is smallest and equal to $50\text{cm}$. The next larger length of column resonating with same tuning fork is

• A. $100\text{cm}$
• B. $125\text{cm}$
• C. $175\text{cm}$
• D. $150\text{cm}$

Answer 1

The correct option is D $150\text{cm}$

We know that for a closed organ pipe, the resonate frequency is given by,
$f=\frac{nv}{4l}$
For length, $l=50\text{cm}=0.5\text{m}$,
frequency,
$f=\frac{nv}{4\times 0.5}$
$\Rightarrow f=\frac{v}{4\times 0.5}$
[length of air column is smallest, so, $n=1$]
To resonate with the same tuning fork at next higher modes, the length of the air column should be increased.
Let us suppose, the new length of the air column is $l^{{}^{\prime }}$, then
$f=\frac{(n+2)v}{4l^{{}^{\prime }}}$
$\Rightarrow f=\frac{(1+2)v}{4l^{{}^{\prime }}}$
So,
$\frac{v}{4\times 0.5}=\frac{(1+2)v}{4l^{{}^{\prime }}}$
On solving this, we get,
$l^{{}^{\prime }}=1.5\text{m}=150\text{cm}$

OR

Given smallest length of column of organ pipe closed at one end and open at another is $50\text{cm}$
$\Rightarrow \frac{\lambda }{4}=50\text{cm}$
hence next mode will be resonating with length of $=3\times \frac{\lambda }{4}=3\times 50\text{cm}=150\text{cm}$