298 problems found
A thin uniform beam \(AB\) has mass \(3m\) and length \(2h\). End \(A\) rests on rough horizontal ground and the beam makes an angle of \(2\beta\) to the vertical, supported by a light inextensible string attached to end \(B\). The coefficient of friction between the beam and the ground at \(A\) is \(\mu\). The string passes over a small frictionless pulley fixed to a point \(C\) which is a distance \(2h\) vertically above \(A\). A particle of mass \(km\), where \(k < 3\), is attached to the other end of the string and hangs freely.
The Fibonacci numbers are defined by \(F_0 = 0\), \(F_1 = 1\) and, for \(n \geqslant 0\), \(F_{n+2} = F_{n+1} + F_n\).
The random variable \(X\) has probability density function \[\mathrm{f}(x) = \begin{cases} kx^n(1-x) & 0 \leqslant x \leqslant 1\,,\\ 0 & \text{otherwise}\,,\end{cases}\] where \(n\) is an integer greater than 1.
Solution:
Two light elastic springs each have natural length \(a\). One end of each spring is attached to a particle \(P\) of weight \(W\). The other ends of the springs are attached to the end-points, \(B\) and \(C\), of a fixed horizontal bar \(BC\) of length \(2a\). The moduli of elasticity of the springs \(PB\) and \(PC\) are \(s_1 W\) and \(s_2 W\) respectively; these values are such that the particle \(P\) hangs in equilibrium with angle \(BPC\) equal to \(90^\circ\).
Solution:
Solution: \begin{questionparts} \item \(\,\) \begin{align*} && I_n &= \int_0^{\beta} (\sec x + \tan x)^n \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} \left ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( (\sec x + \tan x)^{2}+1\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( \sec^2 x + \tan^2 x + 2\sec x \tan x + 1\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( 2\sec x \tan x +2\sec^2 x \right) \, \d x \\ &&& = \left [\frac1n(\sec x + \tan x)^{n} \right]_0^{\beta} \\ &&&= \frac1n[(\sec \beta + \tan \beta)^n - 1] \end{align*} Notice that by AM-GM \(\tfrac12( ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}) \geq (\sec x + \tan x)^{n}\) with equality not holding most of the time. Integrating we obtain our result. \item \(\,\) \begin{align*} && J_n &= \int_0^{\beta} (\sec x \cos \beta + \tan x )^n \d x \\ && \tfrac12( J_{n+1} + J_{n-1}) &= \tfrac12 \int_0^{\beta} \left ( (\sec x \cos \beta + \tan x )^{n+1} +(\sec x \cos \beta + \tan x )^{n-1}\right ) \d x \\ && &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( (\sec x \cos \beta + \tan x )^{2} + \right ) \d x \\ && &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec^2 x \cos^2 \beta + \tan^2 x+ 2\sec x \tan x \cos \beta +1 \right ) \d x \\ && &= \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\tfrac12(\cos^2 \beta +1)\sec^2 x \right ) \d x \\ && &< \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\sec^2 x \right ) \d x \\ &&&= \left [\frac1n (\sec x \cos \beta + \tan x)^{n} \right]_0^{\beta} \\ &&&= \frac1n[ (1 + \tan \beta)^n - \cos^n \beta] \end{align*} But notice we can use the same AM-GM argument from before to show that \(J_n < \tfrac12( J_{n+1} + J_{n-1}) < \frac1n[ (1 + \tan \beta)^n - \cos^n \beta]\)
A sequence \(x_1, x_2, \ldots\) of real numbers is defined by \(x_{n+1} = x_n^2 - 2\) for \(n \geqslant 1\) and \(x_1 = a\).
Solution:
A sequence \(u_1, u_2, \ldots, u_n\) of positive real numbers is said to be unimodal if there is a value \(k\) such that \[u_1 \leqslant u_2 \leqslant \ldots \leqslant u_k\] and \[u_k \geqslant u_{k+1} \geqslant \ldots \geqslant u_n.\] So the sequences \(1, 2, 3, 2, 1\);\ \(1, 2, 3, 4, 5\);\ \(1, 1, 3, 3, 2\) and \(2, 2, 2, 2, 2\) are all unimodal, but \(1, 2, 1, 3, 1\) is not. A sequence \(u_1, u_2, \ldots, u_n\) of positive real numbers is said to have property \(L\) if \(u_{r-1}u_{r+1} \leqslant u_r^2\) for all \(r\) with \(2 \leqslant r \leqslant n-1\).
Solution:
Point \(A\) is a distance \(h\) above ground level and point \(N\) is directly below \(A\) at ground level. Point \(B\) is also at ground level, a distance \(d\) horizontally from \(N\). The angle of elevation of \(A\) from \(B\) is \(\beta\). A particle is projected horizontally from \(A\), with initial speed \(V\). A second particle is projected from \(B\) with speed \(U\) at an acute angle \(\theta\) above the horizontal. The horizontal components of the velocities of the two particles are in opposite directions. The two particles are projected simultaneously, in the vertical plane through \(A\), \(N\) and \(B\). Given that the two particles collide, show that \[d\sin\theta - h\cos\theta = \frac{Vh}{U}\] and also that
The score shown on a biased \(n\)-sided die is represented by the random variable \(X\) which has distribution \(\mathrm{P}(X = i) = \dfrac{1}{n} + \varepsilon_i\) for \(i = 1, 2, \ldots, n\), where not all the \(\varepsilon_i\) are equal to \(0\).
A light elastic spring \(AB\), of natural length \(a\) and modulus of elasticity \(kmg\), hangs vertically with one end \(A\) attached to a fixed point. A particle of mass \(m\) is attached to the other end \(B\). The particle is held at rest so that \(AB > a\) and is released. Find the equation of motion of the particle and deduce that the particle oscillates vertically. If the period of oscillation is \(\dfrac{2\pi}{\Omega}\), show that \(kg = a\Omega^2\). Suppose instead that the particle, still attached to \(B\), lies on a horizontal platform which performs simple harmonic motion vertically with amplitude \(b\) and period \(\dfrac{2\pi}{\omega}\). At the lowest point of its oscillation, the platform is a distance \(h\) below \(A\). Let \(x\) be the distance of the particle above the lowest point of the oscillation of the platform. When the particle is in contact with the platform, show that the upward force on the particle from the platform is \[ mg + m\Omega^2(a + x - h) + m\omega^2(b - x). \] Given that \(\omega < \Omega\), show that, if the particle remains in contact with the platform throughout its motion, \[ h \leqslant a\left(1 + \frac{1}{k}\right) + \frac{\omega^2 b}{\Omega^2}. \] Find the corresponding inequality if \(\omega > \Omega\). Hence show that, if the particle remains in contact with the platform throughout its motion, it is necessary that \[ h \leqslant a\left(1 + \frac{1}{k}\right) + b, \] whatever the value of \(\omega\).
The continuous random variable \(X\) is uniformly distributed on \([a,b]\) where \(0 < a < b\).