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question 5 of 20
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question 5
problem reference 10.1
a 0.450 kg block is attached to an unstrained horizontal spring with spring constant 74.0 n/m. the block is given a displacement of 0.0800 m and then released from rest.
what is the angular frequency ω of the resulting oscillatory motion?
164 rad/s
12.8 rad/s
33.3 rad/s
7.80 × 10⁻² rad/s

Explanation:

Step1: Recall angular frequency formula for spring

$\omega = \sqrt{\frac{k}{m}}$ where $k$=spring constant, $m$=mass

Step2: Substitute given values

$k=74.0$ N/m, $m=0.450$ kg: $\omega = \sqrt{\frac{74.0}{0.450}}$

Step3: Calculate the result

$\sqrt{\frac{74.0}{0.450}} \approx \sqrt{164.44} \approx 12.8$ rad/s

Answer:

B. 12.8 rad/s