1. Resonance condition for a closed pipe

A pipe open at one end and closed at the other supports only odd harmonics:

$f_n = (2n-1)f_0, \qquad n=1,2,3,\dots$

where

$f_0 = \frac{v}{4L}$

is the fundamental frequency.

So resonance occurs if the frequency of sound reaching the pipe is one of:

$f_0,\; 3f_0,\; 5f_0,\dots$

2. Apparent frequency heard at the pipe due to Doppler effect

The source is approaching the stationary pipe with speed $u$. For a moving source approaching a stationary observer, the observed frequency is

$f' = f_s\frac{v}{v-u}$

We must check for each option whether $f'$ equals an odd multiple of $f_0$.

3. Check each option

Option A: $u=0.8v$, $f_s=f_0$

Observed frequency:

$f' = f_0\frac{v}{v-0.8v} = f_0\frac{v}{0.2v} = 5f_0$

Since $5f_0$ is an odd harmonic of a closed pipe, resonance occurs.

✅ A is correct

Option B: $u=0.8v$, $f_s=2f_0$

Observed frequency:

$f' = 2f_0\frac{v}{v-0.8v} = 2f_0\frac{1}{0.2} = 10f_0$

But $10f_0$ is not an odd multiple of $f_0$. So resonance does not occur.

❌ B is incorrect

Option C: $u=0.8v$, $f_s=0.5f_0$

Observed frequency:

$f' = 0.5f_0\frac{v}{v-0.8v} = 0.5f_0\frac{1}{0.2} = 2.5f_0$

Now,

$2.5f_0 = \frac{5}{2}f_0$

which is not of the form $(2n-1)f_0$. Hence no resonance.

❌ C is incorrect

Option D: $u=0.5v$, $f_s=1.5f_0$

Observed frequency:

$f' = 1.5f_0\frac{v}{v-0.5v} = 1.5f_0\frac{1}{0.5} = 3f_0$

Since $3f_0$ is an odd harmonic, resonance occurs.

✅ D is correct

4. Final result

The combinations that produce resonance are:

$\boxed{A \text{ and } D}$

This matches the stored correct answer.