Saved Bookmarks
| 1. |
A stretched string fixed at both end has n nods, then the lengths of the string isA. `(m-1)(lambda)/(2)`B. `((m+1)lambda)/(2)`C. `(mlambda)/(2)`D. `(m -2)(lambda)/(2)` |
|
Answer» Correct Answer - A For p number of loops in a stretched string, the length is given by `l = (plambda)/(2)" "……..(i)` As, number of harmonics = number of loops = number of anti - nodes = P `" "……….(ii)` Also , number of nodes = number of anti - nodes + 1 . Here, number of nodes = m Number of anti - nodes = m - 1 from Eq.(ii) `p = m - 1` Putting this value of p in Eq.(i) we get `l = ((m -1)lambda)/(2)` |
|