A taut string at both ends viberates in its `n^(th)` overtone. The distance between adjacent Node and Antinode is found to be 'd'. If the length of the string is L, then

A taut string fixed at both ends vibrates in n^(th) overtone. The separation between adjacent node and antinode is equal to 'd'. Then string length is

In a pipe open at both the ends the distance between node and antinode is

A string of length L fixed at both ends vibrates in its first overtone. Then the wavelength will be

In stationary wave, the distance between a node and its adjacent antinode is ……….

A string fixed at both ends has consecutive standing wave modes for which the distance between adjacent nodes are 18 cm and 16 cm respectively. The length of the string is -

The string fixed at both ends has standing wave nodes for which distance between adjacent nodes is x_1 . The same string has another standing wave nodes for which distance between adjacent nodes is x_2 . If l is the length of the string then x_2//x_1=l(l+2x_1) . What is the difference in numbers of the loops in the two cases?

A stretched string fixed at both end has n nods, then the lengths of the string is

A string is vibrating in its fifth overtone between two rigid supports 2.4 m apart. The distance between successive node and antinode is

A string is vibrating in its fifth overtone between two rigid supports 2.4 m apart. The distance between successive node and antinode is

A string is vibrating in its fifth overtone between two rigid supports 2.4 m apart. The distance between successive node and antinode is