A sound wave is incident on a pool of fresh water.

The sound enters the water perpendicularly and travels adistance = 0.51 m

before striking a 0.24m mthick copper block lying on the bottom.

The sound passes through the block, reflects from thebottom surface of the block, and returns to the top of the wateralong the same path.

How much time elapses between when the sound enters andleaves the water?

For this problem first we find the time taken by thesound wave to travel through water a distance and down through copper through adistance for this we can apply

\(\displaystyle\triangle{t}\) water=(distancetravelled by sound wave in water)/( speed of sonud in water)ZSK

\(\displaystyle{\frac{{{0.51}\ {m}}}{{{1482}\ \frac{{m}}{{s}}}}}={3.44}\times{10}^{{-{4}}}\)

\(\displaystyle\triangle\)t copper = (distance travelled bysound wave in copper) /( speed of sonud in copper)

\(\displaystyle={\frac{{{0.24}{m}}}{{{5010}\ \frac{{m}}{{s}}}}}={4.79}\times{10}^{{-{5}}}\)

time elapses between the sound enters and leaves thewater can be obtained by applying

\(\displaystyle\triangle{s}\) total=2(\(\displaystyle\triangle{s}{t}\) water+\(\displaystyle\triangle{s}{t}\) copper)

\(\displaystyle={2}{\left[{\left({3.44}\times{10}^{{-{4}}}\right)}+{\left({4.79}\times{10}^{{-{5}}}\right)}\right]}\)

\(\displaystyle={7.84}\times{10}^{{-{4}}}\) s

The sound enters the water perpendicularly and travels adistance = 0.51 m

before striking a 0.24m mthick copper block lying on the bottom.

The sound passes through the block, reflects from thebottom surface of the block, and returns to the top of the wateralong the same path.

How much time elapses between when the sound enters andleaves the water?

For this problem first we find the time taken by thesound wave to travel through water a distance and down through copper through adistance for this we can apply

\(\displaystyle\triangle{t}\) water=(distancetravelled by sound wave in water)/( speed of sonud in water)ZSK

\(\displaystyle{\frac{{{0.51}\ {m}}}{{{1482}\ \frac{{m}}{{s}}}}}={3.44}\times{10}^{{-{4}}}\)

\(\displaystyle\triangle\)t copper = (distance travelled bysound wave in copper) /( speed of sonud in copper)

\(\displaystyle={\frac{{{0.24}{m}}}{{{5010}\ \frac{{m}}{{s}}}}}={4.79}\times{10}^{{-{5}}}\)

time elapses between the sound enters and leaves thewater can be obtained by applying

\(\displaystyle\triangle{s}\) total=2(\(\displaystyle\triangle{s}{t}\) water+\(\displaystyle\triangle{s}{t}\) copper)

\(\displaystyle={2}{\left[{\left({3.44}\times{10}^{{-{4}}}\right)}+{\left({4.79}\times{10}^{{-{5}}}\right)}\right]}\)

\(\displaystyle={7.84}\times{10}^{{-{4}}}\) s