A and B are two fixed points on a smooth horizontal surface, with AB = 3a m. One end of a light elastic string, of natural length a m and modulus of elasticity mg N, is attached to the point A. The other end of this string is attached to a particle P of mass m kg. One end of a second light elastic string, of natural length ka m and modulus of elasticity 2mg N, is attached to B. The other end of this string is attached to P. Given that the system is in equilibrium when P is at M, the mid-point of AB, find the value of k. The particle P is released from rest at a point between A and B where both strings are taut. Show that P performs simple harmonic motion and state the period of the motion. In the case where P is released from rest at a distance 0.2a m from M, the speed of P is 0.7 m s⁻¹ when P is 0.05a m from M. Find the value of a.

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A and B are two fixed points on a smooth horizontal surface, with AB = 3a m - CIE - A-Level Further Maths - Question 3 - 2015 - Paper 1

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A and B are two fixed points on a smooth horizontal surface, with AB = 3a m. One end of a light elastic string, of natural length a m and modulus of elasticity mg N,... show full transcript

Worked Solution & Example Answer:A and B are two fixed points on a smooth horizontal surface, with AB = 3a m - CIE - A-Level Further Maths - Question 3 - 2015 - Paper 1

Step 1

Find the value of k

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Answer

To find the value of k, we start by analyzing the forces acting on the particle P when the system is in equilibrium. The force due to the first string (attached at A) and the second string (attached at B) must be equal:

$mg \frac{(a/2)^2}{ka} = 2mg \frac{(3a/2 - ka)}{ka}$

By simplifying the above equation, we get:

$\frac{mg a}{4k} = 2mg \frac{(3a - 2k)}{2ka}$

Equating the two expressions leads to:

$1 = 6 - 4k$

Thus, we find:

$k = \frac{5}{4}.$

Step 2

Show that P performs simple harmonic motion and state the period of the motion

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Answer

The motion of P can be described by the equation of simple harmonic motion (SHM). Since the restoring force is proportional to the displacement from the equilibrium position, we can express:

$F = -kx$

where k is the effective spring constant. The equation of motion will resemble:

$\frac{d^2x}{dt^2} + \left(\frac{g}{3a}\right)x = 0$

From this, we see that the angular frequency (\omega) is given by:

$\omega = \sqrt{\frac{g}{3a}}.$

The period T of SHM is:

$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{3a}{g}}.$

Step 3

Find the value of a

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Answer

To find the value of a, we analyze the situation at when P is released 0.2a m from M:

The speed is given as 0.7 m s⁻¹ when P is 0.05a m from M.
Using conservation of energy or the equations of motion for SHM:

$v^2 = \omega^2 x$

where (v = 0.7) m/s, (x = 0.05a):

$0.7^2 = \left(\sqrt{\frac{g}{3a}}\right)^2 \cdot 0.05a$

Solving this gives us:

$0.49 = \frac{g \cdot 0.05a}{3}$

Using the gravitational constant g = 9.81:

$0.49 = \frac{9.81 \cdot 0.05a}{3}$

Rearranging yields:

$a = \frac{0.49 \cdot 3}{9.81 \cdot 0.05}.$

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A and B are two fixed points on a smooth horizontal surface, with AB = 3a m - CIE - A-Level Further Maths - Question 3 - 2015 - Paper 1

Worked Solution & Example Answer:A and B are two fixed points on a smooth horizontal surface, with AB = 3a m - CIE - A-Level Further Maths - Question 3 - 2015 - Paper 1

Find the value of k

Show that P performs simple harmonic motion and state the period of the motion

Find the value of a

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