Question

String 1 in the figure has linear density 2.60 g/m and string 2 has linear density 3.30 g/m. A student sends pulses in both directions by quickly pulling up on the knot, then releasing it. She wants both pulses to reach the ends of the strings simultaneously.

What should the string length L1 be?

What should the string length L2 be?

Answer 1

Step-by-step explanation:

We can use the formula for the speed of waves on a string:

v = sqrt(T/μ)

where v is the speed of the wave, T is the tension in the string, and μ is the linear mass density (mass per unit length) of the string.

Let's denote the tension in both strings by T. Since the pulses must reach the ends of both strings simultaneously, we must have:

L1/v1 = L2/v2

where L1 and L2 are the lengths of the strings, v1 is the speed of the wave on string 1, and v2 is the speed of the wave on string 2.

Using the formula above and solving for T, we can eliminate T from this equation to get:

sqrt(μ1/ T)/ L1 = sqrt(μ2/T)/ L2

Squaring both sides and rearranging, we obtain:

L2/L1 = sqrt(μ2/μ1)

Substituting the given values for μ1 and μ2, we get:

L2/L1 = sqrt(3.30/2.60) = 1.126

Solving for one of the lengths, say L1, in terms of the other, we get:

L1 = L2/1.126

Now we need to find the values of L1 and L2 that satisfy the condition that both pulses reach the ends of the strings simultaneously. To do this, we can use the fact that the time it takes for a wave to travel a distance L on a string is given by:

t = L/v

where v is the speed of the wave on the string.

Therefore, if the pulses are to arrive at the ends of the strings simultaneously, we must have:

L1/v1 + L2/v2 = 2L1/v1

Simplifying this equation using the relation L1 = L2/1.126 and the formula for v, we get:

sqrt(T/μ1)L2/1.126/2.60 + sqrt(T/μ2)L2/3.30 = 2L2/1.126sqrt(T/μ1)

Simplifying further and eliminating T, we obtain:

L2 = (2.60/3.30)^2(1.126) L1

Substituting the expression for L1 in terms of L2 that we found earlier, we get:

L2 = (2.60/3.30)^2(1.126) L2/1.126

Solving for L2, we find:

L2 = 2.196 L1

Finally, using the relation L1 = L2/1.126, we get:

L1 = 1.91 m

L2 = 4.20 m

Therefore, the length of string 1 should be 1.91 m and the length of string 2 should be 4.20 m in order for both pulses to reach the ends of the strings simultaneously.

Answer 2

Wave speed on a string is determined by the string's tension and linear mass density, calculated using v = √(T/μ). For two strings with different linear mass densities but equal tension, the time for pulses to pass each other is determined by dividing 2.00 m by each string's wave speed.

The question revolves around the concept of wave speed on a string, which depends on the tension in the string and its linear mass density. The formula to calculate wave speed (v) is v = √(T/μ), where T is the tension in the string and μ is the linear mass density. When a pulse is sent down a string, its speed determines how long it will take to travel to the end of the string.

To find the tension required for a wave speed of 25.00 m/s given an initial speed of 20.00 m/s with a tension of 7.00 N, we can set up the relationship based on the formula for wave speed: v1/v2 = √(T1/T2), where v1 and v2 are the wave speeds and T1 and T2 are the corresponding tensions.

To calculate the time it takes for the pulses to pass one another on two strings with different linear mass densities but the same tension, we utilize the wave speed formula. The length of both strings is the same, so their respective times (t1 and t2) to meet can be found by dividing the distance by their wave speeds: t = 2.00 m / v.