How many nine-letter arrangements can be made using the letters of the word ISOSCELES? A particle moves in a straight line and its position at time t is given by $x = 4 \sin \left( 2\pi t + \frac{\pi}{3} \right)$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2003 - Paper 1

To show that the motion is simple harmonic, we start with the given position function:

[ x = 4 \sin \left( 2\pi t + \frac{\pi}{3} \right). ]

This can be identified as the standard form of simple harmonic motion:

[ x = A \sin(\omega t + \phi), ]

where ( A = 4 ) (the amplitude), ( \omega = 2\pi ) (the angular frequency), and ( \phi = \frac{\pi}{3} ) (the phase constant).

Since the position is described by a sine function, we confirm that the motion is simple harmonic.

The amplitude of a simple harmonic motion is given by the coefficient of the sine function in the equation of motion.

In the equation:

[ x = 4 \sin \left( 2\pi t + \frac{\pi}{3} \right), ]

the amplitude is simply the constant 4. Thus, the amplitude of the motion is 4.

The maximum speed in simple harmonic motion occurs when the particle passes through its equilibrium position (where ( x = 0 )).

Setting the position equation to zero gives:

[ 4 \sin \left( 2\pi t + \frac{\pi}{3} \right) = 0. ]

This implies:

[ \sin \left( 2\pi t + \frac{\pi}{3} \right) = 0 ]

The sine function equals zero at integer multiples of ( \pi ):

[ 2\pi t + \frac{\pi}{3} = n\pi \quad (n \in \mathbb{Z}). ]

Solving for ( t ), we get:

[ t = \frac{n\pi - \frac{\pi}{3}}{2\pi} = \frac{n}{2} - \frac{1}{6}. ]

For the first occurrence, set ( n = 1 ):

[ t = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}. ]

Thus, the particle first reaches maximum speed at ( t = \frac{1}{3} ) seconds.

When two dice are tossed, there are 36 possible outcomes (6 sides on die one and 6 sides on die two). The combinations that yield a sum of 5 include:

• (1, 4)
• (2, 3)
• (3, 2)
• (4, 1)

This gives 4 favorable outcomes. The probability of getting a sum of 5 is then calculated as:

[ P(5) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{36} = \frac{1}{9}. ]

Let ( X ) be the random variable representing the number of times a sum of 5 occurs in 7 tosses. ( X ) follows a binomial distribution with parameters ( n = 7 ) (the number of trials) and ( p = \frac{1}{9} ) (the probability of success).

We need to find ( P(X \geq 2) ) which can be calculated as:

[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1). ]

Using the binomial probability formula:

[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, ]

calculating for ( k = 0 ) and ( k = 1 ):

[ P(X = 0) = \binom{7}{0} \left(\frac{1}{9}\right)^0 \left(\frac{8}{9}\right)^7 = 1 \cdot 1 \cdot \left(\frac{8}{9}\right)^7 ]

[ P(X = 1) = \binom{7}{1} \left(\frac{1}{9}\right)^1 \left(\frac{8}{9}\right)^6 = 7 \cdot \frac{1}{9} \cdot \left(\frac{8}{9}\right)^6. ]

Thus,

[ P(X \geq 2) = 1 - (\frac{8^7}{9^7}) - \left(7 \cdot \frac{1}{9} \cdot \frac{8^6}{9^6}\right). ]

Base Case: For ( n = 1 ):

[ LHS = \frac{1}{1 \times 3} = \frac{1}{3} ] [ RHS = \frac{1}{3}. ]

Thus, the base case holds.

Inductive Step: Assume the formula holds for some ( n = k ), i.e.,

[ \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \ldots + \frac{1}{(2k-1)(2k+1)} = \frac{k}{2k+1}. ]

Now we must show it holds for ( n = k + 1 ):

[ LHS = \frac{k}{2k+1} + \frac{1}{(2(k+1)-1)(2(k+1)+1)}. ]

Substituting: [ = \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)}. ]

Finding common denominators, [ = \frac{k(2k+3) + 1}{(2k+1)(2k+3)} ] [ = \frac{(k+1)}{2k+3}. ]

Thus, this holds true, completing the induction.