1. Mass-Spring Vibration Question
A 410 g block at the end of a spring vibrates 4.3 times per second with an amplitude of 23.0 cm,
as shown in the diagram below. Answer the questions below corresponding to this diagram.
A] Determine the total energy of the system in J.
-total energy of a mass-spring system can be found using the formula 𝐸 =
1
2
𝑘𝑥2
-before we can use this formula to calculate the total energy, we must find k (the spring constant)
-we can do so by using the formula for frequency 𝑓 =
1
2𝜋
√
𝑘
𝑚
-by making the formula solve for the spring constant instead of frequency, we get that 𝑘 = (2𝑓𝜋)2
𝑚
-plugging in the values given by the question, we get that 𝑘 = (2(4.3)𝜋)2
0.41
which solves for: 𝑘 = 299.3
𝑁
𝑚
-we can now use this value in our formula for total energy 𝐸 =
1
2
𝑘𝑥 2
𝐸 =
1
2
(299.3)(0.23)2
𝐸 = 7.92 𝐽
B] Determine the velocity when it passes the equilibrium point in m/s.
-Equilibrium point is when the mass is at its starting point, which is when x=0
-Since velocity is at its maximum when x=0, we are looking for the maximum velocity
-maximum velocity can be found by using the formula 𝐸 =
1
2
𝑚𝑣 2
-by making the formula solve for the velocity instead of energy, we get that 𝑣 = √
2𝐸
𝑚
Xmax = 23.0cmXmin = -23.0cm
Mb = 410g

2. 𝑣 = √
2(7.92)
(0.41)
𝑣 = 6.22 𝑚/𝑠
C] Determine the velocity when it is 0.09 m from equilibrium in m/s.
-the energy kept in the spring when the spring is extended gives us the amount of energy less the energy
of the spring at 0.09m is than the total energy or 𝐸𝑡𝑜𝑡𝑎𝑙 − 𝐸𝑠𝑡𝑜𝑟𝑒𝑑 = 𝐸
-We can use 𝐸 =
1
2
𝑘𝑥 2 to find the stored energy when x = 0.09m
𝐸 =
1
2
(299.3)(0.09)2
𝐸 = 1.21 𝐽
-Now we can subtract the stored energy from the total energy
𝐸𝑡𝑜𝑡𝑎𝑙 − 𝐸𝑠𝑡𝑜𝑟𝑒𝑑 = 𝐸
7.92 − 1.21 = 𝐸
6.71 𝐽 = 𝐸
-Now that we have the energy of the system, we can use 𝐸 =
1
2
𝑚𝑣 2 to find the velocity
6.71 =
1
2
(0.41)𝑣2
(2)(6.71)
0.41
= 𝑣2
5.72 𝑚/𝑠 = 𝑣