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An object on a spring oscillates in simple harmonic motion with frequency f = 1 Hz. If the spring is exchanged for a new one with a spring constant that is half as large as the previous one, what is the new frequency of oscillation?

Respuesta :

Answer:

Explanation:

Given

Frequency of an object in SHM is [tex]f=1\ Hz[/tex]

Frequency in SHM is given by

[tex]f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}[/tex]

where k=spring constant

m=mass of object

if spring is exchanged such that new spring constant is half of previous one then

[tex]k'=0.5 k[/tex]

[tex]f'=\frac{1}{2\pi }\sqrt{\frac{0.5 k}{m}}[/tex]

[tex]f'=\frac{1}{\sqrt{2}}\times \frac{1}{2\pi }\sqrt{\frac{k}{m}}[/tex]

i.e. [tex]f'=\frac{1}{\sqrt{2}}\times 1[/tex]

[tex]f'=\frac{1}{\sqrt{2}}[/tex]

[tex]f'=0.707\ Hz[/tex]