7. A particle \(P\) of mass 0.4 kg is moving in a straight line with simple harmonic motion. The maximum speed of \(P\) is \(5\ \text{m s}^{-1}\) and the maximum magnitude of acceleration of \(P\) is \(12.5\ \text{m s}^{-2}\) The time taken for one complete oscillation is \(T\) seconds.
(a) Show that \(T = \dfrac{4\pi}{5}\) (4)
(b) Find the time within each complete oscillation for which \(P\) is more than 0.5 m from the centre of the oscillation. (4)
(c) Find the kinetic energy of \(P\) at the instant when \(P\) is 0.2 m from the centre of the oscillation. (3)
Mark scheme (a)
Scheme
Marks
AO
\(a\omega = 5\)
B1
3.4
\(a\omega^2 = 12.5\)
B1
3.4
Complete method to find \(T\)
M1
2.1/3.1a
\(\omega = \dfrac{12.5}{5} \Rightarrow T = \dfrac{2\pi}{\omega} = \dfrac{4\pi}{5}\) *
A1 *
1.1b
(4)
Notes
B1: Correct formula for max speed. Accept \(a^2\omega^2 = 25\)
B1: Correct formula for max acceleration. Must be same sign on both sides in their solution.
M1: Complete method e.g. solve for \(\omega\) and use \(T = \dfrac{2\pi}{\omega}\)
A1 *: cso
Mark scheme (b)
Scheme
Marks
AO
\(a = 2 \ \Rightarrow\ x = 2\cos\dfrac{5}{2}t\) or \(x = 2\sin\dfrac{5}{2}t\)
M1
3.4
\(\dfrac{1}{2} = 2\cos\dfrac{5}{2}t \Rightarrow t = \dfrac{2}{5}\cos^{-1}\dfrac{1}{4}\) or \(\dfrac{1}{2} = 2\sin\dfrac{5}{2}t \Rightarrow t = \dfrac{2}{5}\sin^{-1}\dfrac{1}{4}\)
M1
3.1a
Total time: \(4\left(\dfrac{2}{5}\cos^{-1}\dfrac{1}{4}\right)\) or \(\dfrac{4\pi}{5} - 4\left(\dfrac{2}{5}\sin^{-1}\dfrac{1}{4}\right)\)
M1
1.1b
\(= 2.1\) (s)
A1
1.1b
(4)
Notes
M1: Use their \(\omega\) to solve for \(a\) and state or imply equation in \(x\) and \(t\)
M1: Find a relevant value of \(t\)
M1: Find the required value of \(t\) Some candidates will find two times and combine: \(a\sin\omega t = 0.5\ [t_1 = 0.10107,\ t_2 = 1.15556] \rightarrow \text{Time} = 2(t_2 - t_1)\) \(a\cos\omega t = 0.5\ [t_3 = 0.52724,\ t_4 = 0.72939] \rightarrow \text{Time} = 4\pi/5 - 2(t_4 - t_3)\)
5. A particle \(P\) moves in a straight line with simple harmonic motion about a fixed point \(O\). The magnitude of the greatest acceleration of \(P\) is \(18\ \text{m s}^{-2}\)
When \(P\) is 0.3 m from \(O\), the speed of \(P\) is \(2.4\ \text{m s}^{-1}\)
The amplitude of the motion is \(a\) metres.
(a) Show that \(a = 0.5\) (5)
(b) Find the greatest speed of \(P\). (2)
During one oscillation, the speed of \(P\) is at least \(2\ \text{m s}^{-1}\) for \(S\) seconds.
(c) Find the value of \(S\). (4)
Mark scheme (a)
Scheme
Marks
AO
\(a\omega^2 = 18\)
B1
3.4
Use \(v^2 = \omega^2\left(a^2 - x^2\right)\)
M1
3.4
\(2.4^2 = \omega^2\left(a^2 - 0.3^2\right)\)
A1
1.1b
Form equation in \(a\) only and solve for \(a\) e.g. \(\dfrac{a}{a^2 - 0.09} = \dfrac{18 \times 25}{144} \quad \left(18a^2 - 5.76a - 1.62 = 0\right)\)
M1
3.1a
\(\Rightarrow a = 0.5\) *
A1*
1.1b
(5)
Notes
B1: Use the model to state correct equation for greatest acceleration. Accept \(\pm 18\).
M1: Use the model to form a second equation in \(a\) and \(\omega\)
A1: Correct unsimplified equation with \(x\) and \(v\) substituted.
M1: Solve the simultaneous equations to obtain \(a\)
A1*: Correct only from correct working. Must have used positive \(\omega^2\).
Mark scheme (b)
Scheme
Marks
AO
Solve for \(\omega\) and use max speed \(= a\omega\)
B1ft: Set up a correct model to find time when speed is 2. ft their \(\omega\) (If starting from \(O\) then \(|v| = a\omega\cos\omega t\) )
M1: Solves their equation(s) to obtain a critical value for \(t\) \((2 = 3\cos 6t \Rightarrow t = 0.14017...)\) (condone degrees: \(41.8^\circ/48.2^\circ\))
M1: Correct method to obtain \(S\); must be in radians \((S = 4 \times 0.14017...)\)
The fixed points \(A\) and \(B\) lie on a smooth horizontal surface with \(AB = 6\) m.
A particle \(P\) has mass 0.3 kg.
One end of a light elastic string, of natural length 2 m and modulus of elasticity 20 N, is attached to \(P\), and the other end is attached to \(A\).
One end of another light elastic string, of natural length 2 m and modulus of elasticity 40 N, is attached to \(P\) and the other end is attached to \(B\).
The particle \(P\) is at rest in equilibrium at the point \(E\) on the surface, as shown in Figure 7.
(a) Show that \(EB = \dfrac{8}{3}\) m. (3)
The particle \(P\) is now held at the midpoint of \(AB\) and released from rest.
(b) Show that \(P\) oscillates with simple harmonic motion about the point \(E\). (4)
The time between the instant when \(P\) is released and the instant when it first returns to the point \(E\) is \(S\) seconds.
(c) Find the exact value of \(S\). (3)
(d) Find the length of time during one oscillation for which the speed of \(P\) is more than \(2\ \text{m s}^{-1}\) (4)
Mark scheme (a)
Scheme
Marks
AO
In equilibrium: \(T_A = T_B\)
M1
3.3
\(\dfrac{40e}{2} = \dfrac{20(2 - e)}{2}\)
A1
1.1b
\(2e = 2 - e \quad \Rightarrow \quad e = \dfrac{2}{3} \quad \Rightarrow \quad EB = \dfrac{8}{3}\ \text{(m)}\) *
A1*
2.2a
(3)
Notes
M1: Form equation for equilibrium of forces acting on \(P\). Dimensionally correct.
A1: Correct unsimplified equation in one unknown.
A1*: Obtain given answer from correct working. Condone missing units.
Mark scheme (b)
Scheme
Marks
AO
Equation of motion for \(P\) when displaced \(x\) from equilibrium in direction of \(A\)
8.Throughout this question, use \(g = 10\ \text{m s}^{-2}\)
A light elastic string has natural length 1.25 m and modulus of elasticity 25 N.
A particle \(P\) of mass 0.5 kg is attached to one end of the string. The other end of the string is attached to a fixed point \(A\). Particle \(P\) hangs freely in equilibrium with \(P\) vertically below \(A\)
The particle is then pulled vertically down to a point \(B\) and released from rest.
(a) Show that, while the string is taut, \(P\) moves with simple harmonic motion with period \(\dfrac{\pi}{\sqrt{10}}\) seconds. (6)
The maximum kinetic energy of \(P\) during the subsequent motion is 2.5 J.
(b) Show that \(AB = 2\) m (3)
The particle returns to \(B\) for the first time \(T\) seconds after it was released from rest at \(B\)
(c) Find the value of \(T\) (5)
Mark scheme (a)
Scheme
Marks
AO
At equilibrium: \(0.5g = \dfrac{25e}{1.25},\quad e = \dfrac{0.5\times 10\times 1.25}{25} = \dfrac{1}{4}\)
B1
3.3
For taut string, when distance \(x\) from equilibrium, equation of motion
M1
2.1
Alternative for M1: Conservation of energy using a known point (\(E\) or \(B\)) and a general position: From \(E\): \(\dfrac{25e^2}{2\times 1.25} + KE(\text{constant} \ne 0) + 0.5gx = \dfrac{25(e + x)^2}{2\times 1.25} + \dfrac{1}{2}0.5v^2 + 0(\text{GPE})\) and differentiate wrt \(x\) for M1 \(\Rightarrow 0.5g = \dfrac{25(e + x)}{1.25} + \dfrac{1}{2}v\dfrac{\mathrm{d}v}{\mathrm{d}x}\)
B1: Correct only Award if see \(mg = \dfrac{\lambda e}{l}\) used in their equation of motion
M1: Equation of motion with \(x\) measured from the equilibrium position. Need all terms and dimensionally correct. Allow with their \(e \ne 0\). Condone sign errors. Allow with \(a\) in place of \(\ddot{x}\)
A1ft: Correct unsimplified equation with their \(e\) or \(e \ne 0\). Could have the negative of the whole equation or \(x\) replaced with \(-x\) throughout
A1*: Reach given conclusion from correct working. Condone correct conclusion without explanation
M1: Use the model to find the periodic time: \(T = \dfrac{2\pi}{\omega}\) From an equation of the form \(\ddot{x} = -\omega^2 x\)
A1*: Obtain given answer from correct working throughout. Available if only error is not to conclude SHM
Mark scheme (b)
Scheme
Marks
AO
Max KE \(= 2.5 = \dfrac{1}{2}\times\dfrac{1}{2}\times\max v^2 \qquad \Rightarrow \max v^2 = 10\)
B1
1.2
Max speed \(= a\omega\): \(\sqrt{10} = a\sqrt{40}\)
Total time \(= 2\times 0.3311\ldots + \dfrac{2\times\sqrt{7.5}}{10}\)
DM1
3.1a
\(= 1.2\,(\text{s})\) or better
A1
2.2a
(5)
(14 marks)
Notes
B1: Correct equation for SHM seen or implied
M1: Find the time until the string goes slack If working from \(x = 0.5\sin\sqrt{40}t\) need \(\dfrac{T}{4} + \dfrac{1}{\sqrt{40}}\sin^{-1}\dfrac{1}{2}\)
M1: Use the model to find \(v\) or \(v^2\) at the instant the string goes slack \((v = 2.738\ldots)\) Using SHM formula or conservation of energy.
M1: Complete method to find the total time until return to \(B\) Requires the preceding M marks If they use suvat to find the time as a projectile it must be a complete method e.g. \(\sqrt{\dfrac{15}{2}} = -\sqrt{\dfrac{15}{2}} + gt\) or a combination of \(v^2 = u^2 + 2as\) and \(s = ut + \dfrac{1}{2}at^2\)
A1: \(= 1.2\,(\text{s})\) or better Condone an answer to \(\gt\) 2 s.f. Not scored if they have used 9.8.
6. A light elastic string, of natural length \(l\) and modulus of elasticity \(2mg\), has one end attached to a fixed point \(A\) and the other end attached to a particle \(P\) of mass \(m\). The particle \(P\) hangs in equilibrium at the point \(O\).
(a) Show that \(AO = \dfrac{3l}{2}\) (2)
The particle \(P\) is pulled down vertically from \(O\) to the point \(B\), where \(OB = l\), and released from rest.
Air resistance is modelled as being negligible.
Using the model,
(b) prove that \(P\) begins to move with simple harmonic motion about \(O\) with period \(\pi\sqrt{\dfrac{2l}{g}}\) (5)
The particle \(P\) first comes to instantaneous rest at the point \(C\).
Using the model,
(c) find the length \(BC\) in terms of \(l\), (4)
(d) find, in terms of \(l\) and \(g\), the exact time it takes \(P\) to move directly from \(B\) to \(C\). (5)
Mark scheme (a)
Scheme
Marks
AO
\(mg = \dfrac{2mge}{l}\)
M1
3.1a
\(e = \dfrac{1}{2}l\) so \(AO = \dfrac{3l}{2}\) *
A1*
1.1b
(2)
Notes
M1: Use of Hooke’s Law and \(T = mg\)
A1*: Correct answer fully justified
Mark scheme (b)
Scheme
Marks
AO
Equation of motion vertically: \(mg - T = m\ddot{x}\).
M1
2.1
\(mg - \dfrac{2mg(x + e)}{l} = m\ddot{x}\).
A1
1.1b
\(-\dfrac{2g}{l}x = \ddot{x}\), so SHM with \(\omega^2 = \dfrac{2g}{l}\)
M1: All terms needed but allow \(a\) for the acceleration
A1: Correct unsimplified equation including \(\ddot{x}\)
A1: Must mention SHM
M1: Correct method
A1*: Correct answer correctly shown
Mark scheme (c)
Scheme
Marks
AO
Complete method to find \(h\)
M1
3.1a
\(mgh = \dfrac{2mg\left(\dfrac{3l}{2}\right)^2}{2l}\) OR \(v^2 = \dfrac{2g}{l}\left(l^2 - \left(-\dfrac{1}{2}l\right)^2\right)\) and \(0 = \dfrac{3gl}{2} - 2gs\)
A1 A1
1.1b 1.1b
\(h = \dfrac{9l}{4}\)
A1
1.1b
(4)
Notes
M1: Correct no. of terms, dimensionally correct in energy equation OR use SHM to find \(v^2\) at unstretched position AND then use motion under gravity Correct no. of terms, dimensionally correct in both equations
A1: Equation with at most one error
A1: Correct equation OR correct equations
A1: cao
Mark scheme (d)
Scheme
Marks
AO
\(-\dfrac{1}{2}l = l\cos\omega t\)
M1
3.1a
\(t = \dfrac{2\pi}{3}\sqrt{\dfrac{l}{2g}}\)
A1
1.1b
\(v = \sqrt{\dfrac{2g}{l}\left(l^2 - \left(-\dfrac{1}{2}l\right)^2\right)}\) OR
7. A light elastic spring has natural length \(l\) and modulus of elasticity \(4mg\). A particle \(P\) of mass \(m\) is attached to one end of the spring. The other end of the spring is attached to a fixed point \(A\). The point \(B\) is vertically below \(A\) with \(AB = \dfrac{7}{4}l\). The particle \(P\) is released from rest at \(B\).
(a) Show that \(P\) moves with simple harmonic motion with period \(\pi\sqrt{\dfrac{l}{g}}\) (7)
(b) Find, in terms of \(m\), \(l\) and \(g\), the maximum kinetic energy of \(P\) during the motion. (3)
(c) Find the time within each complete oscillation for which the length of the spring is less than \(l\). (5)
Mark scheme (a)
Scheme
Marks
AO
Equation of motion about equilibrium position:
M1
3.1a
\(\dfrac{4mg(x + e)}{l} - mg = -m\ddot{x}\)
A1
1.1b
Extension \(e\) at equilibrium: \(\dfrac{4mge}{l} = mg,\quad \left(e = \dfrac{l}{4}\right)\)
M1: Use the model to find the max speed. Follow their \(\omega\)
M1: Follow their \(a\), \(\omega\)
A1: Correct simplified
Mark scheme (c)
Scheme
Marks
AO
\(x = a\cos\omega t = \dfrac{l}{2}\cos\sqrt{\dfrac{4g}{l}}t\)
B1ft
2.2a
Length of spring \(\lt l \Rightarrow x = -\dfrac{l}{4},\quad -\dfrac{l}{4} = \dfrac{l}{2}\cos\sqrt{\dfrac{4g}{l}}t\)
M1
1.1b
\(\Rightarrow \sqrt{\dfrac{4g}{l}}t = \dfrac{2\pi}{3}\) or \(\dfrac{4\pi}{3}\), \(\quad t = \dfrac{\pi}{3}\sqrt{\dfrac{l}{g}}\) or \(t = \dfrac{2\pi}{3}\sqrt{\dfrac{l}{g}}\)
A1
1.1b
Correct strategy
M1
3.1a
Length of time \(= \dfrac{2\pi}{3}\sqrt{\dfrac{l}{g}} - \dfrac{\pi}{3}\sqrt{\dfrac{l}{g}} = \dfrac{\pi}{3}\sqrt{\dfrac{l}{g}}\)
A1
2.2a
(5)
(15 marks)
Notes
B1: Or equivalent. Follow their \(a\), \(\omega\)
M1: Follow their \(e\) and solve for \(t\)
A1: One correct solution. Accept \(t = \dfrac{2\pi}{3\omega}\), or \(t = \dfrac{4\pi}{3\omega}\)
M1: Complete strategy to find the required interval: select formula for displacement as function of time and use symmetry of motion to find the time interval.
6. The points \(A\) and \(B\) lie on a smooth horizontal surface with \(AB = 4.5\) m.
A light elastic string has natural length 1.5 m and modulus of elasticity 15 N. One end of the string is attached to \(A\) and the other end of the string is attached to \(B\). A particle, \(P\), of mass 0.2 kg, is attached to the stretched string so that \(APB\) is a straight line and \(AP = 1.5\) m. The particle rests in equilibrium on the surface.
The particle is now moved directly towards \(A\) and is held on the surface so \(APB\) is a straight line with \(AP = 1\) m.
The particle is released from rest.
(a) Prove that \(P\) moves with simple harmonic motion. (5)
(b) Find
(i) the maximum speed of \(P\) during the motion,
(ii) the maximum acceleration of \(P\) during the motion. (3)
(c) Find the total time, in each complete oscillation of \(P\), for which the speed of \(P\) is greater than \(5\ \text{m s}^{-1}\). (5)
Mark scheme (a)
Scheme
Marks
AO
Use of Hooke’s law seen
B1
2.1
Let \(x\) be the displacement from equilibrium, then \(T_B - T_A = -0.2\ddot{x}\).
\(\Rightarrow \ddot{x} = -225x = -15^2x\), which is of the form \(\ddot{x} = -\omega^2 x\), so SHM *
A1*
3.2a
(5)
Notes
B1: Use of Hooke’s law seen at least once (with natural length \(\leqslant 1.5\)).
M1: Form equation of motion involving a difference of 2 tensions. Must be dimensionally correct. Need all terms. Condone incorrect lengths. Condone sign errors.
A1: Unsimplified equation with at most one error
A1: Correct unsimplified equation
A1*: All correct and justification of SHM – comment required
Mark scheme (b)
Scheme
Marks
AO
\(a = 0.5,\quad \omega = 15\) or their \(\omega\)
B1ft
1.1b
Max speed \(= a\omega = 7.5\ (\text{m s}^{-1})\)
B1ft
1.2
Max acceleration \(= a\omega^2 = 112.5\ (\text{m s}^{-2})\)
B1ft
1.2
(3)
Notes
B1ft: Correct values seen or implied. \(a\) must be correct but follow their \(\omega\) from (a)
Speed \(\gt 5\) for \(\dfrac{2\pi}{15} - 4t = 0.22\) (s)
A1
1.1b
(5)
(13 marks)
Notes
B1ft: Use correct equation for speed \((-a\omega\sin\omega t)\) Follow their \(a\) and \(\omega\). Allow sin or cos form Can also use correct expression for \(x\) combined with use of \(v^2 = \omega^2\left(a^2 - x^2\right)\)
M1: Find a relevant time e.g. time taken for \(0\ \text{m s}^{-1}\) to \(5\ \text{m s}^{-1}\)
A1: Seen or implied
M1: Complete strategy for the required time – e.g. find value for \(t\) when \(v = 5\) and use symmetry and periodic time
