Elastic Strings and Simple Harmonic Motion

Question

The Ceiling of a hall is 15m above the floor. A vertical elastic string of natural length 5m and modulus of elasticity 6N has one end attached to the ceiling and the other end attached to the floor. A lamp of mass 0.5kg is held at a height of 9m above the floor, and attached to the string. When the lamp is let go, find the heights between which it will oscillate, and the period of the oscillation.

I have tried to do this question using energy conservation, but this is fruitless as is resolving the forces for I do not know:

If the point at which the lamp is attached causes a split in the tension forces, thus tension acts up the string and down it around the lamp.

If this is in fact the case, how do I work out the tension in these separated sections?

Please, someone talk me through a better way to go about this.

Answers - Oscillates between 7.04m and 9m; Period is 1.99s

Answer 1

Consider the string divided into two pieces, so that the natural length of the upper part is 2 and the lower part is 3. First you have to find the position of equilibrium. If the extension in the upper part is $x$ and in the lower part is $y$, then $mg +2y=3x$ and $x+y=10$ which gives $x=4.98$

The mass will perform SHM with centre at the equilibrium position, therefore the amplitude is $0.98$, and therefore the mass oscillates between 9 and 7.04 metres.

Considering a displacement of $x$ from the equilibrium position, we can establish, using Newton's Law and Hooke's Law the equation of motion...

Answer 2

I also have this solution:

The natural length of the upper and lower demarcations are 2 and 3 respectively. Let tension in upper string be T(1) and T(2) for the lower.

Resolving vertically (up as positive) and at equilibrium -

Elastic modulus = 6; Extension in upper string henceforth = (4 + x) and in lower = (6 - x);

T(1) = Weight + T(2)

6(4+x)/2 = 0.5*9.8 + 6(6-x)/3

12 + 3x = 12 - 2x + 4.9

5x = 4.9

Therefore, midpoint of oscillation is when x = 0.98;

This value of 0.98m is also the amplitude of the oscillation.

Resolving vertically (up as +ve) at any time t :

T(1) - mg - T(2) = m * d^2x/dt^2

6(4+x)/2 - 6(6-x)/3 - 4.9 = 0.5 * d^2x/dt^2

12 + 3x - 12 +2x - 4.9 = 0.5 * d^2x/dt^2

5x - 4.9 = 0.5 * d^2x/dt^2

take factor of -5 from LHS

-5(0.98 - x) = 0.5 * d^2x/dt^

d^2x/dt^2 = -10(0.98 - x)

let y = 0.98 - x

d^2y/dt^2 = -10y

we know that n = √k ; d^2y/dt^2 = - ky

Therefore, n = √10

Period of oscillation = 2*pi/n = 1.986... s

Since the amplitude is 0.98, the lamp will fall through a distance (2 x 0.98) = 1.96.

The lamp is dropped from height 9m; it will thus oscillate between 9m and (9-1.96)m = 7.04m

I hope this helps somebody else in the future.