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A string that is 10 cm long is fixed at both ends. At t=0, a pulse travelling from left to right at `1cm//s is 4.0 cm ` from the right end as shown in figure. Determine the next two times when the pulse will be at that point again. State in each case whether the pulse is upright or inverted . . |
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Answer» Correct Answer - B::D `t_(1) = d_(1)/v = (4+4)/1 = 8s`. After one reflection from fixed boundary wave is inverted. `t_2 = d_(2)/v = (4+10+16)/1 = 20s ` After two times reflection from fixed boundaries wave pulse will again become upright. |
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