Correct Answer - Option 1 : 0.5 Hz
CONCEPT:
- Consider a string of length L, stretched under tension T. Let m be the mass per unit length of the string.
The speed of the transverse wave on the string will be:
\(v= \sqrt \frac{T}{m}\)
If in general the string in p segments, then it can be written as:
\(v= \frac{p}{2L}\sqrt \frac{T}{m}\)
CALCULATION:
Here L is 3 m = 300cm, T is 1 kg wt = 1000 × 10 =104 dyne , p = 3 and mass per unit length is given by:
m = 300/300 = 1 g cm-1.
- When the rope vibrates in p segments, its frequency of vibration is given by the formula:
- Where T is tension, L is the length of the string, m is the mass per unit length.
\(v= \frac{3}{2\;\times \;300}\sqrt \frac{10^4}{ 1}=0.5\) Hz
- Hence the frequency of shaking will make it break up into three vibrating segments is 0.5 Hz.
- The first mode of vibration: If the string is plucked in the middle and released, it vibrates in one segments with nodes at its end and an antinode in the middle then the frequency of the first mode of vibration is given by:
\(v= \frac{1}{2L}\sqrt \frac{T}{m}\)
- The second mode of vibration: If the string is plucked in the middle and released, it vibrates in two segments with nodes at its end and an antinode in the middle then the frequency of the second mode of vibration is given by:
\(v= \frac{2}{2L}\sqrt \frac{T}{m}\)
- If in general the string in p segments, then it can be written as:
\(v= \frac{p}{2L}\sqrt \frac{T}{m}\)
