Question

A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string

Answer 1

The time it takes for a pulse to travel a 3.00 m long string with a tension of 500.00 N and a mass of 5.00 g is approximately 5.48 milliseconds.

Step-by-step explanation:

To determine how long it takes a pulse to travel the 3.00 m length of the string, we first need to calculate the wave speed on the string using the formula for wave speed on a stretched string: v = √(T/μ), where v is the wave speed, T is the tension in the string, and μ is the linear mass density of the string.

The linear mass density μ can be calculated by dividing the total mass of the string (m) by its length (L), which gives μ = m/L = 5.00 g / 3.00 m = 0.00167 kg/m (note that we convert the mass to kilograms to keep the units consistent). Now we can calculate the wave speed:

v = √(500.00 N / 0.00167 kg/m) = √(299401.20 m²/s²) ≈ 547.18 m/s.

Finally, to find the time (t) it takes for the pulse to travel the length of the string, we use the relationship 't = L/v', which gives:

t = 3.00 m / 547.18 m/s ≈ 0.00548 s or approximately 5.48 milliseconds.

Answer 2

The time interval is
`t = 5.48 *10^(-3) \ s`

Step-by-step explanation:

From the question we are told that

The length of the string is
`l = 3.00 \ m`

The mass of the string is
`m = 5.00 \ g = 5.0 *10^(-3)\ kg`

The tension on the string is
`T = 500 \ N`

The velocity of the pulse is mathematically represented as

`v = \sqrt{ (T)/(\mu ) }`

Where
`\mu` is the linear density which is mathematically evaluated as

`\mu = (m)/(l)`

substituting values

`\mu = (5.0 *10^(-3))/(3)`

`\mu = 1.67 *10^(-3) \ kg /m`

Thus

`v = \sqrt{(500)/(1.67 *10^(-3)) }`

`v = 547.7 m/s`

The time taken is evaluated as

`t = (d)/(v)`

substituting values

`t = (3)/(547.7)`

`t = 5.48 *10^(-3) \ s`