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You are at a large outdoor concert, seated 300 m from speaker system. The concert is also being broadcast live. Consider a listener 5000 km away who receives the broadcast. Who will hear the music first, you or listener and by what time difference?(Speed of light = 3 × 108 m/s and speed of sound in air = 343 m/s) |
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Answer» s1 = 300 m, v1 = 343 m/s, ∴ t1 = \(\frac{s_1}{v_1}\) = \(\frac{300}{343}\) = 0.8746 s Now, s2 = 5000 km = 5 × 106 m, v2 = c = 3 × 108 m/s ∴ t2 = \(\frac{s_2}{c}\) = \(\frac{5\times10^6}{3\times10^8}\) = 0.0167 s ∴ t2 < t1 ∴ Listener will hear the music first. Time difference = t1 – t2 = 0.8746 – 0.0167 = 0.8579 s The listener will hear the music first, about 0.8579 s before the person present at the concert. |
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