A 100g wire is held under a tension of 250 N with one end at x = 0 and the other x = 10m. At time t = 0, pulse 1 is sent along the wire from the end at x = 10m. At time t = 30ms, pulse 2 is sent along the wire from the x = 0. At what position x do the pulses begin to meet?
The mass per unit length of wire is
s = 0.1 kg/10 m = 10^-2 kg/m
The wave speed is
v = sqrt(T/s) = sqrt[250/10^-2] = 158.1 m/s
With a 30 ms delay of the second pulse, the first pulse will already have traveled x1 = 4.743 m. It will travel half of the remaining distance,
x2 = (1/2)(5.257) = 2.629 m,
before they meeet.
The meet at location x1 + x2.
To determine the position where the pulses begin to meet, we need to consider the speed at which each pulse travels along the wire.
First, we need to calculate the wave speed for the given wire. The wave speed can be determined using the equation:
Wave speed (v) = √(Tension (F) / Linear mass density (μ))
Where:
- Tension (F) is the tension in the wire (250 N in this case)
- Linear mass density (μ) is the mass per unit length of the wire (which can be calculated by dividing the total mass of the wire by its length)
Given that the wire has a mass of 100g (0.1kg) and a length of 10m, we can calculate the linear mass density as follows:
Linear mass density (μ) = Mass (m) / Length (L)
= 0.1kg / 10m
= 0.01 kg/m
Using this value, we can now calculate the wave speed:
Wave speed (v) = √(Tension (F) / Linear mass density (μ))
= √(250 / 0.01)
= √25000
≈ 158.11 m/s
Now that we have the wave speed, we can determine the positions where the pulses begin to meet.
Pulse 1 is sent from the end at x = 10m, and it travels at the wave speed in the negative x-direction. It reaches the position x1 at time t1, which can be calculated using the equation:
x1 = x - v * t
where:
- x is the initial position of the pulse (10m)
- v is the wave speed (-158.11 m/s)
- t is the time elapsed in seconds (which can be calculated by converting 30ms to seconds: 30ms / 1000 = 0.03s)
Plugging in these values, we can find x1:
x1 = 10 - (-158.11 * 0.03)
= 10 + 4.7433
= 14.7433 m
Similarly, Pulse 2 is sent from the end at x = 0, and it travels at the wave speed in the positive x-direction. It reaches the position x2 at time t2. We can use the same equation to calculate x2:
x2 = x + v * t
where:
- x is the initial position of the pulse (0m)
- v is the wave speed (158.11 m/s)
- t is the time elapsed in seconds (0.03s)
Plugging in these values, we can find x2:
x2 = 0 + (158.11 * 0.03)
= 4.7433 m
Since Pulse 1 started from x = 10m and traveled towards the left, while Pulse 2 started from x = 0m and traveled towards the right, at the position where the pulses meet, the distance traveled by each pulse must be equal. Therefore, we can equate x1 and x2:
14.7433 = 4.7433 + x2
Simplifying this equation:
x1 - x2 = 14.7433 - 4.7433
x2 = 10 m
Therefore, the pulses begin to meet at x = 10m.
To determine the position where the pulses begin to meet, we need to calculate the speed at which each pulse travels along the wire.
The speed of a pulse traveling along a wire is given by the equation:
v = sqrt(Ft/μ)
Where:
- v is the velocity or speed of the pulse
- F is the tension in the wire
- t is the time it takes for the pulse to travel the distance
- μ is the linear mass density of the wire, which is equal to the mass per unit length of the wire (μ = m/L)
First, we need to find the linear mass density (μ) of the wire. We can do this by dividing the mass (m) of the wire by its length (L):
μ = m/L
Since we know the mass (m) of the wire is 100g (0.1kg) and the length (L) is 10m, we can calculate:
μ = 0.1kg / 10m
μ = 0.01 kg/m
Next, we can calculate the velocity (v) of each pulse.
For pulse 1 (sent at t = 0 from x = 10m):
- F = 250 N
- t = 0 (since pulse 1 starts at t = 0)
- μ = 0.01 kg/m
v1 = sqrt(250 N * 0 / 0.01 kg/m)
v1 = 0 m/s
For pulse 2 (sent at t = 30ms from x = 0):
- F = 250 N
- t = 30ms (or 0.03s)
- μ = 0.01 kg/m
v2 = sqrt(250 N * 0.03s / 0.01 kg/m)
v2 = sqrt(7500 m^2/s^2)
v2 ≈ 86.6 m/s
Since pulse 1 is not moving (v1 = 0 m/s) and pulse 2 is moving at a velocity of v2 = 86.6 m/s, the pulses will meet when pulse 2 reaches pulse 1.
Using the equation:
v = (Δx) / (Δt)
Where:
- v is the velocity of pulse 2 (86.6 m/s)
- Δx is the position difference between the two pulses (x)
- Δt is the time difference between the two pulses (30ms or 0.03s)
Rearranging the equation to solve for Δx, we have:
Δx = v * Δt
Δx = 86.6 m/s * 0.03s
Δx ≈ 2.6 m
Therefore, the pulses begin to meet at a position x = 2.6 meters from the end of the wire where pulse 2 is sent (x = 0).