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A source of sound of frequency `f_1` is placed on the ground. A detector placed at a height is released from rest on this source. The observed frequency `f(Hz)` is plotted against time `t(sec)` . The speed of sound in air is `300(m)/(s)`. Find `f_1` (`g=10(m)/(s)`).A. `0.5xx10^3Hz`B. `2xx10^3Hz`C. `0.25xx10^3Hz`D. `0.2xx10^3Hz` |
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Answer» `f=((V+V_0)/(V))f_1=f_1+f_1(V_0)/(V)` `V_0="gt"` So, `f+f_1+((f_1g)/(V))t` Slope of graph`=(f_1g)/(V)` `((2)(10^-3)-f_1)/(30)=((f_1)(10))/(300)` or `f_1=10^3Hz` |
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