Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

When a mass of 3.0-kg is hung on a vertical spring, it stretches by 0.085 m. Determine
the period of oscillation of a 4.0-kg object suspended from this spring.

Sagot :

onyebuchinnaji

Answer:

the period of oscillation of the given object is 0.14 s

Explanation:

Given;

mass of the object, m = 3 kg

extension of the spring, x = 0.085 m

The spring constant is calculated as follows;

[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]

The angular speed of a 4 kg object is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]

Therefore, the period of oscillation of the given object is 0.14 s

Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.