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For a closed pipe, such as the one described here, the fundamental frequency (first harmonic) is given by:

f1=v2Lf1=2Lv

Where:

• f1f1 is the fundamental frequency,
• vv is the speed of sound in air (340 m/s in this case),
• LL is the length of the pipe (20 cm = 0.2 m in this case).

Plugging in the values:

f1=3402×0.2=850 Hzf1=2×0.2340=850 Hz

Now, let's find out which harmonic mode of the pipe is resonantly excited by a 430 Hz source. For a closed pipe, the resonant frequencies are odd harmonics. So, we can find the nearest odd multiple of the fundamental frequency to 430 Hz:

fn=(2n−1)f1fn=(2n−1)f1

430 Hz≈(2n−1)×850 Hz430 Hz≈(2n−1)×850 Hz

n≈430850+12n≈850430+21

n≈0.505n≈0.505

Since nn must be an integer for the harmonic mode, we take the nearest integer, which is n=1n=1. So, the first harmonic mode (fundamental frequency) of the pipe is resonantly excited by a 430 Hz source.

Now, let's consider if both ends of the pipe are open. For an open pipe, the fundamental frequency (first harmonic) is given by:

f1=v2Lf1=2Lv

Plugging in the values:

f1=3402×0.2=850 Hzf1=2×0.2340=850 Hz

Now, for an open pipe, the resonant frequencies are all harmonics (both odd and even). So, the same 430 Hz source will not resonate with the pipe because it does not match any of the resonant frequencies.

In conclusion, the harmonic mode of the pipe resonantly excited by a 430 Hz source is the fundamental mode (first harmonic) for a closed pipe. However, the same source will not resonate with the pipe if both ends are open due to the different nature of resonant frequencies for open pipes.