Two stretched strings are placed next to one another, one with a length of 70.0

Question

Two stretched strings are placed next to one another, one with a length of 70.0 cm and the other with a length of 83.0 cm. The two strings have the same mass per unit length. You pluck the shorter one so that it vibrates at its fundamental frequency. You then adjust the tension in the longer string until it resonates with the first one. To resonate, the two strings must have the same fundamental frequency, so that vibrations on one string can cause the second string to vibrate. (a) What is the ratio of the speed of waves on the shorter string to the speed of waves on the longer string?

Explanation / Answer

The fundamental frequency of first string f1= v1/(2× L1)

The fundamental frequency of second string f2= v2/(2x L2)

v1---->speed of wave in shorter string

v2----->speed of wave in longer string

L1------->length of first string

L2-------->lengtg of second string

As f1= f2.,

v1/(2× L1)= v2/(2×L2)

v1/v2 = L2/L1

=1.1857

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