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A 0.50-kg object is attached to a horizontal spring whose spring constant is k=300 N/m and undergoing a simple harmonic motion. Calculate its (a) Period (b) frequency (c) angular frequency

User Jozef Izso
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ANSWER -

We can use the following formulas to calculate the period, frequency, and angular frequency of an object undergoing simple harmonic motion:

(a) Period (T) = 2π * √(m/k), where m is the mass of the object and k is the spring constant.

(b) Frequency (f) = 1/T, where T is the period.

(c) Angular frequency (ω) = 2π * f = 2π / T

Given the mass of the object (m) = 0.50 kg and the spring constant (k) = 300 N/m, we can calculate:

(a) Period (T) = 2π * √(m/k) = 2π * √(0.50 kg / 300 N/m) = 0.69 seconds.

(b) Frequency (f) = 1/T = 1/0.69 s = 1.45 Hz.

(c) Angular frequency (ω) = 2π * f = 2π / T = 2π / 0.69 s = 9.07 rad/s.

Therefore, the period of the motion is 0.69 seconds, the frequency is 1.45 Hz, and the angular frequency is 9.07 rad/s.
User Jarmond
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