Q.A girl stands 170 m away from a high wall and claps her hands at a steady rate so that each clap coincides with the echo of the one before. a) If she makes 60 claps in 1 minute, what value should be the speed of sound in air? b) Now, she moves to another location and finds that she should now make 45 claps in 1 minute to coincide with successive echoes. Calculate her distance for the new position from the wall.

[!TLDR] a) v = 2(170)/1 = 340 m/s from the 60-claps-per-minute (1 s interval) round trip. b) Using the same v with the new 45-claps-per-minute (4/3 s) interval gives a working distance of about 226.7 m, though the book prints 255 m for this part. [!ANSWER] a) v = 340 …

Part a) 60 claps in 1 minute means the interval between successive claps is $T_1=\dfrac{60\text{ s}}{60}=1$ s. Since each clap is timed to coincide with the ECHO of the PREVIOUS clap, this 1 s interval must equal the round-trip travel time of sound to the wall (170 m away) and back: $2d_1=vT_1\Rightarrow v=\dfrac{2d_1}{T_1}=\dfrac{2\times170}{1}=340$ m/s. Part b) At the new location, 45 claps in 1 minute gives an interval $T_2=\dfrac{60}{45}=\dfrac{4}{3}\approx1.333$ s. Using the SAME speed of sound (340 m/s, just established), the round-trip condition gives the new distance $d_2$: $2d_2=vT_2\Rightarrow d_2=\dfrac{vT_2}{2}=\dfrac{340\times\frac{4}{3}}{2}=\dfrac{453.33}{2}\approx226.7$ m. Note on the printed answer: the textbook's own printed answer key states b) 255 m, but substituting the given numbers (170 m, 60 claps/min, 45 claps/min) through the same round-trip relation used correctly in part (a) gives 226.7 m, not 255 m; 255 m would require a clap interval of 1.5 s ( …

Use the round-trip condition (clap interval = 2×distance/speed of sound) separately for each part: solve part (a) for v using the known 170 m distance, th …

Forgetting that 'claps per minute' must first be converted to a TIME INTERVAL between successive claps (interval = 60/rate) before it can be equated to the round-trip travel time; also, …