Two identical mass-spring systems consist of 250-g masses attached to springs of spring constant 15 N/m. If one of them starts oscillating from its equilibrium position at time t = 0, how much later should the second one start so they oscillate out of phase by 90 o?
T = 2pi sqrt(m/k)
Start the second at T/2
To find out how much later the second mass-spring system should start to oscillate out of phase by 90 degrees, we need to consider the formula for the period of an oscillating mass-spring system:
T = 2π √(m / k)
Where:
T = period of oscillation (time taken for one complete oscillation)
m = mass of the object attached to the spring
k = spring constant
Since the two mass-spring systems are identical, they have the same mass and spring constant. Therefore, the period of oscillation for both systems will be the same.
Let's calculate the period of oscillation for one of the systems:
m = 250 g = 0.25 kg (converting grams to kilograms)
k = 15 N/m
T = 2π √(0.25 / 15)
T ≈ 1.137 seconds (rounded to three decimal places)
Now, if we want the second mass-spring system to start oscillating 90 degrees out of phase, we need to find a time delay that is one-fourth of the period of oscillation, since 90 degrees is one-fourth of a complete oscillation.
Time delay = T / 4
Time delay ≈ 1.137 / 4
Time delay ≈ 0.284 seconds (rounded to three decimal places)
Therefore, the second mass-spring system should start oscillating about 0.284 seconds later than the first system in order to be out of phase by 90 degrees.