[" 42.A string,fixed at both ends,vibrates in a resonant mode "],[" with a separation of "2.0cm" between the consecutive "],[" nodes.For the next higher resonant frequency,this "],[" separation is reduced to "1*6cm" .Find the length of the "],[" string."]

A uniform string fixed at both ends vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For next higher resonant frequency, this separation is reduced to 1.6 cm. The length of the string is

A uniform string fixed at both ends vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For next higher resonant frequency, this separation is reduced to 1.6 cm. The length of the string is

A string fixed at both ends vibrates in a resonant mode with a separation of 2.0cm between the consecutive nodes. For the next higher resonants frequecy this separation to 1.6cm. The length of the string is

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

If a string fixed at both ends, vibrates in its fourth harmonic, the wavelength is 15 cm. What is the length of the string ?

If a string fixed at both ends vibrates in four loops. The wavelength is 10 cm. The length of string is

A string fixed at both the ends forms standing waves with node separation of 5 cm. If the velocity of waves travelling in the string is 4 m/s, then the frequency of vibration of the string will be

[" A stretched string of "],[" length "2m" is found to "],[" vibrate in resonance "],[" with a tuning fork of "],[" frequency "420Hz" ."],[" The next higher "],[" frequency for which "],[" resonance occurs is "],[490Hz" .The velocity "],[" of the transverse "],[" wave along this string "]

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 -