Question

A block attached to a spring with unknown spring constant oscillates with a period of 8.0 ss . Parts a to d are independent questions, each referring to the initial situation. What is the period if Part A The mass is doubled

Answer 1

Answer:

What is the period if Part A The mass is doubled

T = √2 * 8 = 11.313 s

Explanation:

The equation for the period of an object attached to a spring is express as:

T=2π$\sqrt{m/k}$=

Where m is the mass and k is the spring constant

If the mass is doubled:

T = √2 * 8 = 11.313 s

Answer 2

The new time period will be 11.312 secons

The time period (T) of the oscillation in a spring with spring constant k, attached to an object of mass m is:

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

T = 8 seconds (given)

Now if the mass is doubled, $m^{'}=2m$

then time period $T^'$:

$T^{'}=2\pi \sqrt{\frac{m^'}{k} } \\\\T^{'}=2\pi \sqrt{\frac{2m}{k} } \\\\T^'}=\sqrt{2}T\\\\T^'}=1.414*8\\\\T^'}=11.312$seconds will be the new time period.

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