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A phone cord is 4.89 m long. The cord has a mass of 0.212 kg. A transverse wave pulse is produced by plucking one end of the taut cord. That pulse makes four round trips (down and back) along the cord in 0.617 s. What is the tension (N) in the cord? Hint: Four round trips is a total distance of eight lengths of the cord. (Answer between 100-300 N)

Answer :

Khoso123

Answer:

Tension in Cord=174 N

Explanation:

Given Data

L (Phone Cord Length)=4.89 m

m (Cord Mass)=0.212 Kg

T (Time for four trips)=0.617 s

Tension=?

Solution

V=λ×f

[tex]V=\frac{8*4.89}{0.617}\\ V=63.4m/s[/tex]

[tex]Sigma=\frac{mass}{length}\\ Sigma=\frac{0.212}{4.89}\\ Sigma=0.0433 \frac{kg}{m}[/tex]

[tex]Wave Speed=\sqrt{\frac{Tension}{Sigma} }\\ \\V=\sqrt{\frac{T}{Sigma} }\\ V^{2}=\frac{T}{Sigma}\\  T=V^{2}*Sigma\\ T=(63.4)^{2}*(0.0433)\\ T=174 N[/tex]

onyebuchinnaji

The tension in the given  phone cord is 174.3 N.

Speed of the wave

The speed of the wave is calculated as follows;

v = fλ

v = (n/t) x 2L

v = (4/0.617) x (2 x 4.89)

v = 63.4 m/s

Tension on the rope

The tension in the rope is calculate as follows;

v = √T/μ

Where;

μ is mass per unit length = 0.212/4.89 = 0.0434 kg/m

v² = T/μ

T = v²μ

T = (63.4)² X 0.0434

T = 174.3 N

Thus, the tension in the given  phone cord is 174.3 N.

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