An unknown mass is hung from a very light spring with a spring constant of 13.2 N/m. Neglect the mass of the spring. If the period of oscillation of the mass-spring system is 2.60 s, then the unknown mass is kg.
Amplitude is the maximum displacement of the mass from the _.
end point equilibrium point top point vibrational point
A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of +14.8 cm at time, t = 0.00 s. The displacement from equilibrium of the mass at time, t = 1.25 s is cm.
A ball of mass 1.55 kg is hung from a spring which stretches a distance of 0.3800 m. The spring constant of the spring (a scalar quantity) is N/m
A mass m on a spring with a spring constant k is in simple harmonic motion with a period T. If the same mass is hung from a spring with a spring constant of 3k, the period of the motion will be _.
increased by a factor of the square root of 3 decreased by a factor of the square root of 3 decreased by a factor of 3 increased by a factor of 3

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Xema8 Xema8 · Answer 1 · 2019-12-04T14:20:20+01:00

Answer:

Explanation:

Time period of oscillation of spring mass system

T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]

m = [tex]\frac{T^2\times k}{4\pi^2}[/tex]

= [tex]\frac{(2.6)^2\times 13.2}{4\pi^2}[/tex]

= 2.26 kg

Amplitude is the maximum displacement of the mass from the _equilibrium point .

maximum displacement from equilibrium of +14.8 cm at time, t = 0.00 s

x(t) = 14.8cosωt

ω = 2πn = 2π x 9.73 = 61.1 rad /s

x(t) = 14.8cos61.1 t

when t = 1.25

x(t) = 14.8cos61.1 x 1.25

= 14.8cos61.1 x 1.25

= 14.8cos76.38

= 8.22 cm

A ball of mass 1.55 kg is hung from a spring which stretches a distance of 0.3800 m

spring constant

k = mg / x

= 1.55 x 9.8 / .38

= 40 N / m

T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]

so if mass is hung from a spring with a spring constant of 3k

T decreases by a factor of the square root of 3