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(D) The linear mass density of the string is $\mu = \frac{m}{L} = \frac{10^{-3} 	ext{ kg}}{0.5 	ext{ m}} = 2 	imes 10^{-3} 	ext{ kg/m}$.The veloci

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$0.05$

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(D) The linear mass density of the string is $\mu = \frac{m}{L} = \frac{10^{-3} 	ext{ kg}}{0.5 	ext{ m}} = 2 	imes 10^{-3} 	ext{ kg/m}$.The velocity of the wave is $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{0.8}{2 	imes 10^{-3}}} = \sqrt{400} = 20 	ext{ m/s}$.The time taken for a pulse to travel the length of the string is $t = \frac{L}{v} = \frac{0.5 	ext{ m}}{20 	ext{ m/s}} = 0.025 	ext{ s}$.For constructive interference to occur between successive pulses, the first pulse must reflect from the fixed end and return to the starting point before the next pulse is generated.Therefore, the minimum time interval is $\Delta t = 2t = 2 	imes 0.025 	ext{ s} = 0.05 	ext{ s}$.

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