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A particle D of mass m is suspended from a fixed point by a light elastic string of natural length ℓ and elastic constant \( \frac{3mg}{ℓ} \) - Leaving Cert Applied Maths - Question 6 - 2021
Question 6
A particle D of mass m is suspended from a fixed point by a light elastic string of natural length ℓ and elastic constant \( \frac{3mg}{ℓ} \). Initially D rests in e... show full transcript
Worked Solution & Example Answer:A particle D of mass m is suspended from a fixed point by a light elastic string of natural length ℓ and elastic constant \( \frac{3mg}{ℓ} \) - Leaving Cert Applied Maths - Question 6 - 2021
Step 1
Show that D moves with simple harmonic motion.
Answer
To demonstrate that D moves with simple harmonic motion (SHM), we start by analyzing the forces acting on the particle when it is displaced downwards by a distance ( x ).
At equilibrium, the tension in the elastic string equals the weight of the mass:
[ T_0 = mg ]
The elastic force exerted by the string can be defined using Hooke's law:
[ ke = mg ]
Where ( k = \frac{3mg}{ℓ} ), thus:
[ e = \frac{x}{ℓ} ]
This indicates that when the particle moves a distance ( x ), the tension becomes:
[ T = mg - \frac{3mg}{ℓ} e = mg - \frac{3mg}{ℓ} \left( \frac{x}{ℓ} \right) = mg - \frac{3mgx}{ℓ^2} ]
By equating the forces and using Newton's second law, we have:
[ ma = -\frac{3mg}{ℓ^2} x ]
Hence, the acceleration is proportional to ( -x ), indicating SHM.
The angular frequency can also be derived:
[ \omega = \sqrt{\frac{3g}{ℓ}} ]
Step 2
In terms of ℓ, find the height above the equilibrium position to which D rises.
Answer
To find the height above the equilibrium position, we first calculate the maximum extension of the elastic string when the kinetic energy is maximum and potential energy is minimum:
When mass D rises, it reaches a point where all potential energy converts to elastic potential energy:
[ 0 = mg(h - \frac{3}{2}ℓ) + \frac{1}{2} k e^2 ]
This simplifies to [ 0 = mg(h - \frac{3}{2}ℓ) + \frac{1}{2} \left( \frac{3mg}{ℓ} \right) \left(\frac{3}{2}ℓ \right)^2 ]
Simplifying gives:
[ 0 = mg(h - \frac{3}{2}ℓ) + \frac{27mg}{8} \rightarrow h = \frac{27}{8}ℓ - \frac{3}{2}ℓ = \frac{27 - 12}{8}ℓ = \frac{15}{8}ℓ ]
Thus, the height above equilibrium is ( \frac{15}{8}ℓ - \frac{3}{2}ℓ = \frac{15 - 12}{8}ℓ = \frac{3}{8}ℓ ).
Step 3
Find the value of \( \theta \).
Answer
Using energy conservation principles for the child moving down the slide, we have:
[ \frac{1}{2} mv^2 = mg(h + (r - r \cos \theta)) ]
Setting ( v^2 = 2g(r - r \cos \theta) \rightarrow v^2 = 2gr(1 - \cos \theta) )
This leads to:
[ mg \cos \theta = \frac{mv^2}{r} = 2mg(1 - \cos \theta) ]
Thus:
[ \cos \theta = \frac{1}{4} \rightarrow \theta = \cos^{-1}(\frac{1}{4}) \approx 75.5^{\circ} ]
Step 4
Find, in terms of r, the speed of the child at K.
Answer
To find the child's speed at K after losing contact, we utilize conservation of mechanical energy:
Considering the change in height from E to K:
[ \frac{1}{2} mv^2 = mg(r + h) - mg ]
This means to derive speed at point K:
[ v^2 = 2g((r + h) - 0) \rightarrow v^2 = 2g(r + \frac{12}{5}r) = 2gr(\frac{17}{5}) ]
Thus, the speed ( v_K = \sqrt{\frac{34gr}{5}} \approx 2.61\sqrt{g} ]