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sleet_krkn [62]
2 years ago
8

A spring stretches 3.5 cm when a 9 g object is hung from it. The object is replaced with a block of mass 26 g which oscillates i

n simple harmonic motion. Calculate the period of motion. The acceleration of gravity is 9.8 m/s 2.
Physics
1 answer:
evablogger [386]2 years ago
3 0

Answer:

The period of motion  of new mass T = 0.637 sec

Explanation:

Given data

Mass of object (m) = 9 gm = 0.009 kg

Δx = 3.5 cm = 0.035 m

We know that spring force is given by

F = m g

F = 0.009 × 9.81 = 0.08829 N

Spring constant

k = \frac{F}{x}

k = \frac{0.08829}{0.035}

k = 2.522 \frac{N}{m}

New mass(m_1) = 26 gm = 0.026 kg

Now the period of motion is given by

T = 2 \pi \sqrt{\frac{m}{k} }

T = 2 \pi \sqrt{\frac{0.026}{2.522} }

T = 0.637 sec

This is the period of motion  of new mass.

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A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic
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<em>A = 6.9 cm</em>

Explanation:

<u>Simple Harmonic Motion</u>

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

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