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The maximum speed of a simple harmonic oscillator is 4 m/s. If the amplitude is
The maximum speed of a simple harmonic oscillator is 4 m/s. If the amplitude is 2 m, what is the angular frequency? (2019)
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Practice Questions
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Q1
The maximum speed of a simple harmonic oscillator is 4 m/s. If the amplitude is 2 m, what is the angular frequency? (2019)
2 rad/s
4 rad/s
1 rad/s
8 rad/s
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Maximum speed (v_max) = ωA => ω = v_max / A = 4 m/s / 2 m = 2 rad/s.
Questions & Step-by-step Solutions
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Q
Q: The maximum speed of a simple harmonic oscillator is 4 m/s. If the amplitude is 2 m, what is the angular frequency? (2019)
Solution:
Maximum speed (v_max) = ωA => ω = v_max / A = 4 m/s / 2 m = 2 rad/s.
Steps: 5
Show Steps
Step 1: Understand the terms. The maximum speed (v_max) of the oscillator is given as 4 m/s, and the amplitude (A) is given as 2 m.
Step 2: Recall the formula that relates maximum speed, angular frequency (ω), and amplitude: v_max = ωA.
Step 3: Rearrange the formula to solve for angular frequency (ω): ω = v_max / A.
Step 4: Substitute the values into the formula: ω = 4 m/s / 2 m.
Step 5: Perform the division: ω = 2 rad/s.
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