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 particle is undergoing simple harmonic motion (SHM), we start with the given equation:

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

1. Form of the equation: The equation is in the form of (x = A\sin(\omega t + \phi)), where:

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

2. Characteristics of SHM: In SHM, the displacement varies sinusoidally with time, confirming that the motion is periodic and the restoring force is proportional to the displacement.

3. Acceleration: The acceleration is given by: $a = -A\omega^2 \sin(\omega t + \phi)$ which is also proportional to the displacement, confirming SHM.

Thus, the particle is indeed undergoing simple harmonic motion.

The amplitude of the motion is the coefficient of the sine function in the equation:

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

Thus, the amplitude is given by:

[ A = 4 ]\

Therefore, the amplitude of the motion is 4 units.

To find when the particle first reaches maximum speed, we note that maximum speed occurs when the derivative of the displacement, the velocity, is at its peak. The velocity is given by:

$v = \frac{dx}{dt} = 4\omega\cos(\omega t + \phi)$

Substituting (\omega = 2\pi) and (\phi = \frac{\pi}{3}):

[ v = 8\pi\cos\left(2\pi t + \frac{\pi}{3}\right) ]

Setting (v = 0) to find maximum speed leads to:

[ \cos\left(2\pi t + \frac{\pi}{3}\right) = 1 ]

This occurs when:

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

Solving for the first instance at (n = 0):

[ 2\pi t = 2\pi - \frac{\pi}{3}
t = \frac{5}{6} \quad \text{(first instance after } t = 0) ]

Thus, the particle first reaches maximum speed at (t = \frac{5}{6}) seconds.

When tossing a pair of dice, there are a total of 36 possible outcomes (6 faces on the first die multiplied by 6 faces on the second die). The ways to achieve a sum of 5 are as follows:

• (1, 4)
• (2, 3)
• (3, 2)
• (4, 1)

This gives us a total of 4 ways to roll a sum of 5. Thus, the probability is calculated as:

[ P(\text{sum} = 5) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{4}{36} = \frac{1}{9} ]

Consequently, the probability of getting a sum of 5 when one pair of fair dice is tossed is indeed (\frac{1}{9}).

To find the probability of obtaining a sum of 5 at least twice in 7 tosses:

1. Determine PMF and Binomial Distribution: We can model this scenario using a binomial distribution where:

   • (n = 7) (number of trials),
   • (p = \frac{1}{9}) (probability of success for each trial).

2. Calculate cumulative probability: We need to calculate:

   • (P(X \geq 2) = 1 - P(X = 0) - P(X = 1)) where (X) is the number of successes (sums of 5).

3. Using the binomial formula:

   • For (P(X = k) = {n \choose k} p^k (1 - p)^{n - k}):
   • Calculate for k = 0 and k = 1:
     • (P(X = 0) = {7 \choose 0} \left(\frac{1}{9}\right)^0 \left(\frac{8}{9}\right)^7 = (\frac{8}{9})^7 )
     • (P(X = 1) = {7 \choose 1} \left(\frac{1}{9}\right)^1 \left(\frac{8}{9}\right)^6 = 7 \cdot \left(\frac{1}{9}\right) \cdot \left(\frac{8}{9}\right)^6)

4. Combine probabilities and calculate:

   • Use complementary probabilities:
   • Finally, compute: $P(X \geq 2) = 1 - P(X = 0) - P(X = 1).$ This will give you the desired probability.

To prove this by induction:

1. Base Case: For (n = 1):

   • Left side: (\frac{1}{1 \times 3} = \frac{1}{3})
   • Right side: (\frac{1}{2 \cdot 1 + 1} = \frac{1}{3})
   • Thus, base case holds true.

2. Inductive Step: Assume true for (n = k):

   • That is: $\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \cdots + \frac{1}{(2k-1)(2k+1)} = \frac{k}{2k + 1}$

3. Prove for (n = k+1):

   • Add the next term: $\frac{1}{(2k+1)(2k+3)}$
   • Therefore: $\frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)}$
   • Combine terms:
   • After algebraic manipulation, you will arrive at: $= \frac{k + 1}{2k + 3}$
   • Hence, it holds that the equation is valid for (n = k + 1).

4. This concludes the inductive proof ensuring that it is true for all positive integers (n).