A mass M = 0.464 kg moves with an initial speed v = 2.98 m/s on a level frictionless air track. The mass is initially a distance D = 0.250 m away from a spring with k = 844 N/m, which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance d, before reversing direction. After bouncing off the spring, the mass travels with the same speed v, but in the opposite direction.
a) Find the total elapsed time until the mass returns to its starting point. (Hint: The mass undergoes a partial cycle of simple harmonic motion while in contact with the spring.)
T=2•π•sqrt(m/k) = 2•π•sqrt(0.464/844)=0.147 s.
The time of “to and fro” motion is T/2 ≈ 0.074 s.
The time of uniform motion before and
after the contact with the spring
t= 2•(D/v) =2•(0.250/2.98)=0.084 s.
The total time is 0.074+ 0.084 =0.158 s.
not the right answer
What is the right answer? I'd like to understand.
no
To find the total elapsed time until the mass returns to its starting point, we need to calculate the time taken for two parts of the motion: when the mass compresses the spring and when it moves back to its starting point after bouncing off the spring.
Let's start by finding the compression distance d of the spring using the potential energy equation for a spring:
Potential Energy (PE) = 0.5 * k * d^2
Where k is the spring constant (844 N/m) and d is the compression distance.
We know that the initial potential energy of the mass is equal to its initial kinetic energy:
0.5 * M * v^2 = 0.5 * k * d^2
Substituting the given values, we have:
0.5 * 0.464 kg * (2.98 m/s)^2 = 0.5 * 844 N/m * d^2
Simplifying the equation:
0.5 * 0.464 kg * (2.98 m/s)^2 = 422 * d^2
Now, solve for d:
d^2 = (0.464 kg * (2.98 m/s)^2) / 422
d^2 = 0.970798
Taking the square root of both sides, we get:
d ≈ 0.985 m
Now let's calculate the time taken by the mass when it's in contact with the spring. We can use the equation for the period of a mass-spring system:
T = 2π * √(m/k)
Where T is the period, m is the mass, and k is the spring constant.
Substituting the given values, we have:
T = 2π * √(0.464 kg / 844 N/m)
T ≈ 2π * √(0.000549 m)
Simplifying the equation:
T ≈ 2π * 0.0234 s
T ≈ 0.1469 s
So, the mass spends approximately 0.1469 seconds in contact with the spring.
Now, let's find the time taken by the mass to travel back to its starting point after bouncing off the spring. Since the speed and direction of the mass remain unchanged, this motion is just a linear motion.
We can use the equation for linear motion to calculate the time:
Time = Distance / Speed
The distance is twice the compression distance (2d) because the mass has to travel the distance it compressed the spring in both directions.
Time = (2 * 0.985 m) / 2.98 m/s
Time ≈ 0.660 s
Therefore, the total elapsed time until the mass returns to its starting point is approximately 0.1469 seconds (time spent in contact with the spring) plus 0.660 seconds (time to travel back to the starting point) equals 0.8069 seconds.