Question

A 0.49-kg cord is stretched between two supports, 7.8m apart. When one support is struck by a hammer, a transverse wave travels down the cord and reaches the other support in 0.83s . What is the tension in the cord?

Answer 1

To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,

$v = \sqrt{\frac{T}{\mu}}$

Here,

v = Velocity

$\mu$= Linear density (Mass per  unit length)

T = Tension

Rearranging to find the Period we have that

$T = v^2 \mu$

$T = v^2 (\frac{m}{L})$

As we know that speed is equivalent to displacement in a unit of time, we will have to

$T = (\frac{L}{t}) ^2(\frac{m}{L})$

$T = (\frac{7.8}{0.83})^2 (\frac{0.49}{7.8})$

$T = 5.54N$

Therefore the tension is 5.54N