Maths Further ( Mechanics)

(1) To calculate the length AB, we can use the fact that the object is in equilibrium, which means that the tension in the string is equal to the weight of the object. The weight of the object is given by its mass multiplied by the acceleration due to gravity, which is approximately 9.81 m/s^2. Therefore, we have:

T = mg

where T is the tension in the string, m is the mass of the object, and g is the acceleration due to gravity. Substituting the given values, we get:

T = (3 kg)(9.81 m/s^2) = 29.43 N

The tension in the string is also given by Hooke’s law:

T = lx

where l is the modulus of elasticity of the string and x is its extension. Since the string is in equilibrium, its extension must be equal to AB - 0.5 m, where AB is the length of the string. Substituting these values into Hooke’s law and equating it with T, we get:

lx = 29.43 N 49x = 29.43 N x = 0.6 m

Therefore, AB = 0.5 m + 0.6 m = 1.1 m.

(ii) To find the tension in each of the strings in terms of x, we can consider each string separately. Let T1 be the tension in the first string and T2 be the tension in the second string.

For string 1:

T1 = lx

For string 2:

T2 = l(xm + 2.5) - mg

Since both strings are identical, they have the same modulus of elasticity l. Substituting x = 0.6 m and xm = 1.4 m (as shown below), we get:

T1 = (49 N/m)(0.6 m) = 29.4 N T2 = (49 N/m)(1.4 m + 2.5 m) - (3 kg)(9.81 m/s^2) = 68.7 N

To show that x = 1.4 m, we can use a similar argument as before: since both strings are identical and in equilibrium, their extensions must be equal to each other.

Therefore:

AB - 0.5 m = xm + 2.5 m 1.1 m - 0.5 m = xm + 2.5 m xm = 1.4 m

(iii) The total elastic potential energy in the strings when the object hangs in equilibrium can be calculated using:

Elastic potential energy stored in string = l x^2 / 2

The total elastic potential energy is simply twice this value since there are two identical strings:

Total elastic potential energy = l x^2

Substituting x = 0.6 m and l = 49 N/m, we get:

Total elastic potential energy = (49 N/m)(0.6 m)^2 = 17.64 J

(iv) When the object is pulled down by 0.1 m from its equilibrium position and released from rest, it oscillates about its equilibrium position with simple harmonic motion.

The period of oscillation can be calculated using:

Period T = 2π√(m/k)

where k is the spring constant and is equal to l/A, where A is the cross-sectional area of the string.

The mass per unit length of a string can be calculated using:

Mass per unit length μ = ρA

where ρ is the density of the string.

Substituting these values into T, we get:

T = 2π√(m/μl)

Substituting x = AB - 0.6 m (the extension when it’s pulled down by 0.1m), we get:

l(x + 0.1) - mg = Tension l(x + 0.1) - (3 kg)(9.81 m/s^2) = μlA(x + 0.1) l(x + 0.1) - (3 kg)(9.81 m/s^2) / A(x + 0.1) / ρA= x + 0.1 l(x + 0.1) - (3 kg)(9.81 m/s^2) / ρA(x + 0.1)^2= x + 0.1 x^3 + (3/7)x^2 - (7/245)x - (343/24500) ≈ -0.0008









For part iii aren't the extensions on either side of the string different? Also, not too sure about how to do part iv) using changes in elastic potential energy and kinetic energy as I got v = 0.40 m/s. Thanks......