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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 \( \frac{6mg}{\ell} \) - Leaving Cert Applied Maths - Question 6 - 2020
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 \( \frac{6mg}{\ell} \). \nInitially D rests... 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 \( \frac{6mg}{\ell} \) - Leaving Cert Applied Maths - Question 6 - 2020
Step 1
(i) Show that D moves with simple harmonic motion.
Answer
To establish that D moves with simple harmonic motion (SHM), we begin by analyzing the forces acting on the particle when it is displaced.\n\nAt the displacement ( x = \frac{1}{2} \ell ), the restoring force ( F_r ) can be defined as the net force acting on the mass due to gravity and the elastic force of the string:
[ F_r = mg - \frac{6mg}{\ell} \cdot x ]
From Hooke's law, the elastic force is given by the extension of the spring. The particle undergoes SHM if the acceleration is proportional to the displacement from the equilibrium position, hence:
[ m\frac{d^2x}{dt^2} = -k x ]
where the effective spring constant ( k = \frac{6mg}{\ell} ). This shows that:
[ \frac{d^2x}{dt^2} + \frac{6g}{\ell} x = 0 ]
which represents simple harmonic motion characterized by an angular frequency ( \omega = \sqrt{\frac{6g}{\ell}} ).
Step 2
(ii) In terms of ℓ, find the greatest speed of D while the string remains taut.
Answer
To find the greatest speed of D while the string remains taut, we apply energy conservation principles. The total mechanical energy in the system is conserved:
Initial potential energy at the maximum displacement (when D is pulled down): [ PE_i = mg\left(\frac{1}{2} \ell \right) = \frac{mg\ell}{2} ]
At the point where the string is taut, all this potential energy is converted to kinetic energy (assuming no energy losses): [ KE = \frac{1}{2}mv^2 ]
Setting the potential energy equal to kinetic energy gives: [ mg\left(\frac{1}{2} \ell \right) = \frac{1}{2}mv^2 ] [ v^2 = g \ell ] [ v = \sqrt{g \ell} ]
Thus, the greatest speed of D while the string remains taut is given in terms of the length ( \ell ): [ v_{max} = \sqrt{g \ell} ] and simplifies to ( \frac{\sqrt{6g}}{3} ) by considering the maximal displacement.
Step 3
(i) Show that \( \cos \theta = \frac{1}{3} \).
Answer
To show that ( \cos \theta = \frac{1}{3} ), we can analyze the forces acting on particle P as it moves in a vertical circle. \n\nAt point B, the tension in the string plus the weight component must equal the centripetal force: [ T + mg \cos \theta = \frac{m v^2}{r} ]\n Since P is projected with speed ( v = \sqrt{3g} ), substituting into the centripetal force equation gives: [ T + mg \cos \theta =\frac{m (3g)}{d/2} ]\n This establishes a relationship between forces, leading us to derive: [T + mg \cdot \frac{1}{3} = mg\text{ (at slack)}] yields ( \cos \theta = \frac{1}{3} ).
Step 4
(ii) In terms of d, find the greatest height of P above B in the subsequent motion.
Answer
For the subsequent motion of particle P, after the string goes slack, we first identify energy conservation principles. As particle P moves upward, we note the kinetic energy at point A (just before slack) converts to gravitational potential energy as it ascends: [ \frac{1}{2} m (\sqrt{3g})^2 = m g h ]
Substituting for kinetic energy: [ \frac{3mg}{2} = mgh ] [ h = \frac{3d}{2g} ]
This denotes the height above B achieved by particle P in the vertical circle as it swings upward to its highest point, given in terms of d.