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A small mass attached to a spring oscillates in simple harmonic motion with an amplitude of . It takes seconds to complete oscillations. Calculate its period, angular velocity, maximum speed and the maximum acceleration.
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Q. A small mass attached to a spring oscillates in simple harmonic motion with an amplitude of . It takes seconds to complete oscillations. Calculate its period, angular velocity, maximum speed and the maximum acceleration.
- Calculate Period: To find the period () of the oscillation, we need to divide the total time taken by the number of oscillations.Total time for oscillations = secondsNumber of oscillations = Period () = Total time for oscillations / Number of oscillations
- Calculate Angular Velocity: Calculating the period:
- Find Maximum Speed: Next, we calculate the angular velocity (), which is given by the formula .
- Calculate Maximum Acceleration: Calculating the angular velocity: $\omega = \frac{\(2\)\pi}{T} = \frac{\(2\)\pi}{\(3\)} \text{ seconds} \approx \(2\).\(0944\) \text{ radians per second}
- Calculate Maximum Acceleration: Calculating the angular velocity:\(\newline\)\(\omega = \frac{2\pi}{T} = \frac{2\pi}{3} \) seconds \(\approx 2.0944\) radians per secondTo find the maximum speed (\(v_{\text{max}}\)), we use the formula \(v_{\text{max}} = \omega \times A\), where \(A\) is the amplitude.\(\newline\)Amplitude (\(A\)) = \(50\)mm = \(0.05\)m (since we need to work in SI units)
- Calculate Maximum Acceleration: Calculating the angular velocity: \(\omega = \frac{2\pi}{T} = \frac{2\pi}{3} \text{ seconds} \approx 2.0944 \text{ radians per second}\)To find the maximum speed \((v_{\text{max}})\), we use the formula \(v_{\text{max}} = \omega \times A\), where \(A\) is the amplitude. Amplitude \((A) = 50\text{mm} = 0.05\text{m}\) (since we need to work in SI units)Calculating the maximum speed: \(v_{\text{max}} = \omega \times A = 2.0944 \text{ radians/second} \times 0.05\text{m} \approx 0.10472 \text{ meters per second}\)
- Calculate Maximum Acceleration: Calculating the angular velocity: \(\omega = \frac{2\pi}{T} = \frac{2\pi}{3} \text{ seconds} \approx 2.0944 \text{ radians per second}\)To find the maximum speed \((v_{\text{max}})\), we use the formula \(v_{\text{max}} = \omega \times A\), where \(A\) is the amplitude. Amplitude \((A) = 50\text{mm} = 0.05\text{m}\) (since we need to work in SI units)Calculating the maximum speed: \(v_{\text{max}} = \omega \times A = 2.0944 \text{ radians/second} \times 0.05\text{m} \approx 0.10472 \text{ meters per second}\)Finally, to find the maximum acceleration \((a_{\text{max}})\), we use the formula \(a_{\text{max}} = \omega^2 \times A\).
- Calculate Maximum Acceleration: Calculating the angular velocity: \(\omega = \frac{2\pi}{T} = \frac{2\pi}{3} \text{ seconds} \approx 2.0944 \text{ radians per second}\)To find the maximum speed \((v_{\text{max}})\), we use the formula \(v_{\text{max}} = \omega \times A\), where \(A\) is the amplitude. Amplitude \((A) = 50\text{mm} = 0.05\text{m}\) (since we need to work in SI units)Calculating the maximum speed: \(v_{\text{max}} = \omega \times A = 2.0944 \text{ radians/second} \times 0.05\text{m} \approx 0.10472 \text{ meters per second}\)Finally, to find the maximum acceleration \((a_{\text{max}})\), we use the formula \(a_{\text{max}} = \omega^2 \times A\). Calculating the maximum acceleration: \(a_{\text{max}} = \omega^2 \times A = (2.0944 \text{ radians/second})^2 \times 0.05\text{m} \approx 0.2197 \text{ meters per second squared}\)
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