Question 266: Waves

Question

A transverse displacement of vibrating string is $y=0.06 \sin \left(\frac{2 \pi}{3}ight) 	imes \cos (120 \pi t)$.
If the mass per unit length of a string is $4 	imes 10^{-2} \mathrm{~kg} / \mathrm{m}$, then the tension in the string will be

Accepted Answer

WeÃ¢ÂÂre tasked with finding the
tension in the vibrating string
.
Step 1: Write the wave equation
The transverse displacement is given as:
$ y = 0.06 \sin\left(\frac{2\pi}{3} xight) \cos(120\pi t) $
(This seems to be implied; otherwise the argument of sine looks incomplete. Likely the full form is $ y = 0.06 \sin(kx) \cos(\omega t)$.)
So here:
Amplitude: $A = 0.06 \, 	ext{m}$ (not needed directly)
$k = 	frac{2\pi}{3} \,	ext{rad/m}$
$\omega = 120 \pi \,	ext{rad/s}$
Step 2: Relating tension, mass per unit length, and wave parameters
For a wave on a string:
$ \omega^2 = v^2 k^2 $
where wave speed $v = \sqrt{	frac{T}{\mu}}$,
with $\mu$ = mass per unit length.
Step 3: Solve for wave speed
$ v = \frac{\omega}{k} $
Plugging values:
$ \omega = 120\pi, \quad k = \frac{2\pi}{3} $
$ v = \frac{120\pi}{	frac{2\pi}{3}} = \frac{120\pi \cdot 3}{2\pi} = 180 \,	ext{m/s}. $
Step 4: Find tension
$ v = \sqrt{\frac{T}{\mu}} \quad \Rightarrow \quad T = \mu v^2 $
Given mass per unit length:
$ \mu = 4 	imes 10^{-2} \,	ext{kg/m} = 0.04. $
So
$ T = 0.04 	imes (180)^2 $
$ T = 0.04 	imes 32400 = 1296 \, 	ext{N}. $
Ã¢ÂÂ Final Answer:
$ 	ext{Tension } = 1296 \, 	ext{N} $
Correct option: A (1296 N).
Ã¢ÂÂ

Answer

1296 N

Answer

1000 N

Answer

720 N

Answer

500 N

Waves

A transverse displacement of vibrating string is y=0.06 \sin \left(\frac{2 \pi}{3}\right) \times \cos (120 \pi t). If the mass per unit length of a string is 4 \times 10^{-2} \mathrm{~kg} / \mathrm{m}, then the tension in the string will be

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