A mass of 0.34 kg is attached to a spring and is set into vibration with a period of 0.16 s. What is the spring constant of the spring? Answer in units of N/m.

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The natural frequency of a mass-spring system is given as:

`f_n = 1/(2pi) sqrt(k/m)`

where, fn is the natural frequency, k is the spring constant and m is the mass.

Here, m = 0.34 kg and time period = 0.16 s

Frequency, fn = 1/time period = 1/0.16 = 6.25...

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The natural frequency of a mass-spring system is given as:

`f_n = 1/(2pi) sqrt(k/m)`

where, fn is the natural frequency, k is the spring constant and m is the mass.

Here, m = 0.34 kg and time period = 0.16 s

Frequency, fn = 1/time period = 1/0.16 = 6.25 hz

Substituting these values in the equation, we get:

`6.25 = 1/(2pi) sqrt(k/0.34)`

`2pi xx 6.25 = sqrt(k/0.34)`

`(2pi xx 6.25)^2 = k/0.34`

`k = 0.34 xx (2pi xx 6.25)^2`

`k = 524.32 N/m`

Thus, the given spring has a spring constant of 524.32 N/m.

Thus, given the mass of the object and time period, we have calculated the spring constant of the spring. Given the spring constant and say mass, we could have calculated the time period of the mass-spring system. Thus, if we are given with any 2 of the 3 parameters (spring constant, mass and time period), we can calculate the missing parameter value.

Hope this helps. 

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