2. Questions 2a, 2b and 2c refer to a 0.250 kg pendulum. (a) What length of the pendulum is needed to oscillate at the same frequency as the object in question 1? Show your work [2 marks] (b) What is the restoring force on the pendulum at an angle of 6.24c from the equilibrium position?Show your work. [2 marks] (c) The pendulum is pulled aside until it is 0.386 m above its lowest position and released. The pendulum is designed to emit sound waves at a frequency of 440 Hz; however, as it swings toward and away from an observer, the frequency appears to vary slightly . What is the apparent frequency of the sound from the pendulum as it swings at its maximum speed toward an observer?Assume the speed of sound is 345m/s Show your work. [4 marks] (d) Regarding 2c, what is this phenomenon called? [1 mark]

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2. Questions 2a, 2b and 2c refer to a 0.250 kg pendulum. (a) What length of the pendulum is needed to oscillate at the same frequency as the object in question 1? Show your work [2 marks] (b) What is the restoring force on the pendulum at an angle of 6.24c from the equilibrium position?Show your work. [2 marks] (c) The pendulum is pulled aside until it is 0.386 m above its lowest position and released. The pendulum is designed to emit sound waves at a frequency of 440 Hz; however, as it swings toward and away from an observer, the frequency appears to vary slightly . What is the apparent frequency of the sound from the pendulum as it swings at its maximum speed toward an observer?Assume the speed of sound is 345m/s Show your work. [4 marks] (d) Regarding 2c, what is this phenomenon called? [1 mark]

Answer 1

## Part (a): With \(g = 9.81 \mathrm{m/s^2}\) and \(f = 0.607 \mathrm{Hz}\), we compute: ### \(L = \frac{g}{(2\pi f)^2} = 6.69 \mathrm{m}\) ## Part (b): With \(m = 0.250 \mathrm{kg}\), \(g = 9.81 \mathrm{m/s^2}\), and \(\theta = 6.24^{\circ}\), we compute: ### \(F = mg \sin{\theta} = 0.027 \mathrm{N}\) ## Part (c): With \(f = 440 \mathrm{Hz}\), \(v = 345 \mathrm{m/s}\), and \(v_0 = 0 \mathrm{m/s}\) (since the pendulum is at its maximum speed and hence momentarily stationary), we compute: ### \(f' = f \frac{v + v_0}{v} = 440 \mathrm{Hz}\) ## Part (d): The phenomenon described in part (c) is known as the Doppler effect.

Answer 2

## Step 1: First, we need to determine the length of the pendulum that will oscillate at the same frequency as the object in question 1. The frequency of the object in question 1 is \(0.607 \mathrm{Hz}\). We can use the formula for the period of a pendulum, which is \(T = 2\pi \sqrt{\frac{L}{g}}\), where \(T\) is the period, \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity. Since the frequency is the reciprocal of the period, we can rearrange the formula to solve for \(L\): ### \(L = \frac{g}{(2\pi f)^2}\) ## Step 2: Next, we need to find the restoring force on the pendulum when it is at an angle of \(6.24^{\circ}\) from the equilibrium position. The restoring force is given by the formula \(F = mg \sin{\theta}\), where \(m\) is the mass of the pendulum, \(g\) is the acceleration due to gravity, and \(\theta\) is the angle from the equilibrium position. ## Step 3: The pendulum is pulled aside until it is \(0.386 \mathrm{m}\) above its lowest position and released. The pendulum is designed to emit sound waves at a frequency of \(440 \mathrm{Hz}\). We need to find the apparent frequency of the sound from the pendulum as it swings at its maximum speed toward an observer. We can use the Doppler effect formula for sound, which is ### \(f' = f \frac{v + v_0}{v}\) where \(f'\) is the apparent frequency, \(f\) is the source frequency, \(v\) is the speed of sound, and \(v_0\) is the speed of the source. ## Step 4: Finally, we need to identify the phenomenon described in part (c). This is known as the Doppler effect.

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