PHYSICS
JAMB 2022 - Question 11
Physics 2022 JAMB Past Questions - Question 11: An object of mass 50g is suspended from the end of a spiral spring of force constant 0.5N/m, the body is set into simple harmonic motion with 0.3m displacement. The period of the motion is
Choose the correct answers from the options given.
A:
B:
C:
D:
Correct Answer
B
Explanation
To find the period of the motion, we can use the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Where:
T = period of the motion
m = mass of the object (in kg)
k = force constant of the spring (in N/m)
First, we need to convert the mass from grams to kilograms:
\[ m = 50g = 0.05kg \]
Now we can use the formula:
\[ T = 2\pi \sqrt{\frac{0.05}{0.5}} \]
\[ T = 2\pi \sqrt{0.1} \]
\[ T = 2\pi \times 0.316 \]
\[ T = 0.632\pi \]
So, the period of the motion is approximately 1.986 seconds.
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