In a lab, a student measures the unstretched length of a spring as 12.9 cm. When a 113.6-g mass is hung from the spring, its length is 26.3 cm. The mass-spring system is set into oscillatory motion, and the student observes that the amplitude of the oscillation decreases by about a factor of 2 after five complete cycles.

a) Calculate the period of oscillation for this system, assuming no damping.
7.34×10-1 s
b) What would be the difference between the period with no damping and the period with damping?
1.79×10-4 s

Can anyone please explain how to solve these problems

mass = .1136 kg

g = 9.81 m/s^2
so
weight = .1136*9.81 = 1.114 Newtons

spring constant
k = 1.114 N/ .134 meters
= 8.32 N/m

T = 2 pi sqrt(m/k)
= 2 pi sqrt (.1136/8.32)
= 0.734 s = 7.34 * 10^-1 s

if damping proportional to v

F = -kx -bv

then x = A e^-(bt/2m) sin w' t

five cycles is about 3.67 seconds
so
1/2 = e^-(3.67 b /.226)
take ln both sides
-.693 = - -16.2 b
so
b = .0427

then w' = sqrt (k/m -b^2/4m^2)
= sqrt (73. - .035)
=8.542 = 2 pi/T
T = .7355 s
change in T = .0015 s =1.5*10^-4 s

You could do this more accurately either carrying more significant figures or using calculus to get dT/db

To solve these problems, we can use the formulas for the period of oscillation and the damping factor in a mass-spring system.

a) To calculate the period of oscillation for this system, assuming no damping, we can use the equation:

T = 2π√(m/k)

Where T is the period, m is the mass, and k is the spring constant.

In this case, we are given the mass of the object (113.6 g) and the unstretched length of the spring (12.9 cm). However, we need to convert the mass to kilograms and the length to meters to match the SI units.

So, first, we convert the mass:
m = 113.6 g = 0.1136 kg

Next, we convert the length:
l_unstretched = 12.9 cm = 0.129 m

Now, we need to find the spring constant, k. The spring constant represents the stiffness of the spring and can be calculated using Hooke's Law:

k = (F_applied) / (l_stretched - l_unstretched)

Where F_applied is the force applied to the spring (weight of the mass) and l_stretched is the length of the spring when the mass is hung.

In this case, F_applied = mg, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

F_applied = (0.1136 kg) * (9.8 m/s^2) = 1.11328 N

l_stretched = 26.3 cm = 0.263 m

Now we can calculate the spring constant:

k = (1.11328 N) / (0.263 m - 0.129 m) = 10.0473 N/m

Finally, we can use the equation for the period of oscillation:

T = 2π√(m/k)
T = 2π√(0.1136 kg / 10.0473 N/m)
T ≈ 7.34×10^(-1) s

So the period of oscillation for this system, assuming no damping, is approximately 7.34×10^(-1) s.

b) To find the difference between the period with no damping and the period with damping, we need to consider the decay of amplitude over time. The amplitude decreases by a factor of 2 after five complete cycles. This indicates that the damping factor, γ, is related to the period, T, by the equation:

γ = (1 / 5T)

Now, we can calculate the difference between the period with no damping (T) and the period with damping (T_damping):

T_damping = T - γ
T_damping = T - (1 / 5T)

Using the value of T calculated in part a:

T_damping ≈ 7.34×10^(-1) s - (1 / (5 * 7.34×10^(-1) s))
T_damping ≈ 7.34×10^(-1) s - 2.71×10^(-1) s
T_damping ≈ 5.63×10^(-1) s

Therefore, the difference between the period with no damping and the period with damping is approximately 1.79×10^(-4) s.

Remember to check the calculations and units to ensure accuracy.