**(New tips are continually
added to these pages. Check back in a few months' time for more)**

**TOPIC
11: Sound**

**Tip 1:**

Echo problems are
teachers' favourites in exams because they are usually more challenging.
Remember to multiply by a factor of '**2**'
to the distance between the source and the reflecting wall, in order to get the
distance traveled by the sound wave.

Example: I stand between two opposite-facing walls separated by 425m and shout. The time interval between hearing the first echo and second echo is 1.5 seconds. How far am I standing from each wall?

Solution: Let *x* be the
distance from me to the nearer wall; and *y* the distance from me to the further
wall. Therefore *x* + *y* = 425m

The distance travelled by
the first echo was **2***x*
(towards the wall and then back to me), so the time it took was **
2***x*/340 seconds (speed of sound is taken as 340 m s^{-1}). Similarly,
the distance traveled by the second echo was **
2***y*,
so the time it took was **2***y*/340
seconds. The time interval (1.5 seconds) was the difference between
**2***y*/340
and **2***x*/340.
Therefore **2***y*/340
- **2***x*/340 = 1.5

Equation
1: *x* + *y* = 425

Equation 2:
**2***y*/340
- **2***x*/340 = 1.5

Two equations, two
unknowns. Solving by simultaneous equation, we get *x* = 85m and *y* = 340m.

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