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A string that is 10 cm long is fixed at ...

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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Text Solution

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The correct Answer is:
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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DC PANDEY ENGLISH-SUPERPOSITION OF WAVES-Level 1 Subjective
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