Physics · Principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, fundamental mode and harmonics, beats
A man standing unsymmetrically between two parallel cliffs, claps his hand and s
A man standing unsymmetrically between two parallel cliffs, claps his hand and starts hearing a series of echoes at intervals of 1 s. If the speed of sound in air is \( 340 \mathrm{ms}^{-1} \), then the distance between the two prallel cliffs, is
- A. \( 170 \mathrm{m} \)
- B. 340 \( \mathrm{m} \)
- C. 510 \mathrm{m} \)
- D. \( 680 \mathrm{m} \)
Step-by-step solution
Let distances to near and far cliffs be d1 and d2. The first echo from near cliff arrives at t1=2d1/v, second from far cliff at t2=2d2/v. The third echo is from double reflection (near→far→man) at t3=2(d1+d2)/v. Intervals: t2-t1=2(d2-d1)/v and t3-t2=2d1/v. Given intervals are 1 s, so 2d1/v=1 and 2(d2-d1)/v=1. Solving: d1=170 m, d2=340 m. Total distance D=d1+d2=510 m.