An empty cart, tied between two ideal springs, oscillates with ω = 10.1 rad/s.A load is placed in the cart, making the total mass 3.9 times what it was before. What is the new value of ω?
rad/s
This is how I did it, but its wrong
3.9/3=1.3 then 1.3x10.1=13.13 rad/s
I don't know what I am doing wrong.
ω1 =sqrt (k/m)
ω2 =sqrt (k/(3.9•m))
ω2/ ω1 = sqrt(k•m/(3.9•m•k)=0.506
ω2 = 0.506•ω1=5.11 rad/s
To solve this problem, you need to apply the concept of conservation of energy in a simple harmonic motion.
1. In an oscillating system, the angular frequency (ω) is related to the mass and the spring constant (k) by the equation ω = √(k/m), where m is the mass and k is the spring constant.
2. Initially, the cart is empty, so let's assume the mass of the cart is m1. Therefore, the original angular frequency is ω1 = √(k/m1).
3. After adding the load, the total mass becomes 3.9 times the original mass, so the new mass is m2 = 3.9m1.
4. We need to find the new angular frequency, ω2. Using the formula ω2 = √(k/m2), we can substitute m2 with 3.9m1 to get ω2 = √(k/(3.9m1)).
5. Now, let's compare ω1 and ω2. We have ω2 = √(k/(3.9m1)) and ω1 = √(k/m1).
6. To find ω2, we need to find the ratio of ω2 to ω1. We can do this by dividing ω2 by ω1.
ω2/ω1 = (√(k/(3.9m1))) / (√(k/m1))
= √(k/(3.9m1)) * √(m1/k)
= √(m1/m1)*(k/k)/(√(3.9))
= √(1/3.9)
≈ 0.507
7. Finally, we calculate the new angular frequency ω2 by multiplying this ratio with the original angular frequency ω1.
ω2 = ω1 * (ω2/ω1)
= 10.1 rad/s * 0.507
≈ 5.12 rad/s
Therefore, the new value of ω is approximately 5.12 rad/s.