Searches Related to 1. A mass of 20 kg is suspended at the end of a rubber cord of diameter 2.0 mm and length 100 cm. Find the period of vertical oscillations of the mass. [Take young’s modulus E, for the rubber = 2.0 x 108 Nm - 2
r = .001 meter
Area = pi r^2 = 3.14 * 10^-6 m^2
delta L/L = F /( E A )
delta L/1 m = F /(2*10*8 *3.14*10^-6)
F/delta L = spring constant = k 628 N/m
now I am sure you can do the sqrt (k/m) thing
I donot understand a thing..
To find the period of vertical oscillations of the mass, we need to consider the properties of the rubber cord, such as its Young's modulus and the physical characteristics of the system.
The period of oscillation is determined by the equation:
T = 2π√(m/k)
where T is the period, m is the mass, and k is the spring constant.
In this case, the rubber cord acts as a spring with a certain spring constant k. The spring constant of the rubber cord can be calculated using Hooke's Law:
F = kx
where F is the force applied to the cord, k is the spring constant, and x is the extension or compression of the cord.
The force applied is the weight of the mass, given by:
F = mg
where m is the mass and g is the acceleration due to gravity.
To calculate the extension or compression of the rubber cord, we can use the formula for elongation in a uniform stretch:
ΔL = (F * L) / (A * E)
where ΔL is the change in length, F is the force applied, L is the length of the cord, A is the cross-sectional area of the cord, and E is the Young's modulus of the rubber.
Now, we can substitute the relevant values into the formulas and calculate the period of vertical oscillations.
Given:
Mass (m) = 20 kg
Diameter of cord (d) = 2.0 mm
Length of cord (L) = 100 cm
Young's modulus (E) = 2.0 x 10^8 Nm^(-2)
First, let's calculate the cross-sectional area (A) of the cord:
A = πr^2
= π(d/2)^2
= π(2.0 mm/2)^2
= π(1.0 mm)^2
≈ 3.14 mm^2
Next, let's calculate the force applied (F) to the cord:
F = m * g
= 20 kg * 9.8 m/s^2
= 196 N
Now, let's calculate the change in length (ΔL) of the cord:
ΔL = (F * L) / (A * E)
= (196 N * 100 cm) / (3.14 mm^2 * 2.0 x 10^8 Nm^(-2))
≈ (196 N * 100 cm) / (3.14 * 0.314 mm^2 x 10^8 Nm^(-2))
≈ (196 N * 100 cm) / (0.314 mm^2 x 10^8 Nm^(-2))
≈ (196 N * 100 cm) / (0.314 x 10^(-4)m^2)
≈ (196 N * 100 x 10^(-2) m) / (0.314 x 10^(-4)m^2)
≈ (196 N * 10^(-1) m) / (0.314 x 10^(-4) m)
≈ (196 N * 10^(-1) m) / (3.14 x 10^(-6) m)
≈ 6.24 x 10^6 m
Finally, let's calculate the period (T) of oscillations:
T = 2π√(m/k)
= 2π√(20 kg / (F/ΔL))
= 2π√(20 kg / ((196 N) / (6.24 x 10^6 m)))
= 2π√(20 kg * (6.24 x 10^6 m) / 196 N)
= 2π√(20 * 6.24 x 10^6 kg * m / 196 N)
= 2π√(6.24 x 10^7 kg * m / N)
Evaluating the above expression will give us the desired period of vertical oscillations of the mass.