Wednesday

March 4, 2015

March 4, 2015

Posted by **sand** on Tuesday, January 10, 2012 at 1:56am.

- physics -
**drwls**, Tuesday, January 10, 2012 at 7:14amUse the formula for natural frequency f:

f = [1/(2 pi)]*sqrt(k/m)

Rearrange it to solve for the spring constant k, knowing frequency (f) and mass (m).

4 pi^2 * f^2 * m = k

The amplitude does not matter in this case. The units of the answer (k) will be Newtons per meter if f is in Hz and m is in kg, since

kg * s^-2 = (kg*m/s^2)/m = N/m

- physics -
**sand**, Tuesday, January 10, 2012 at 8:13amI had already used the above formula and calculated k = 315.8 N/m

But the book answer is k = 101 N/m

Hence this query

- physics -
**drwls**, Tuesday, January 10, 2012 at 9:03amThe book answer appears to be wrong by a factor of pi

- physics -
**sand**, Tuesday, January 10, 2012 at 10:42amThanks a lot

**Answer this Question**

**Related Questions**

physics - A 0.5 Kg mass is supported by a SPRING . The system is set in ...

Physics HELP!!!!!!!! - When an object of mass m1 is hung on a vertical spring ...

physics - A spring with a spring constant of 1.33 102 N/m is attached to a 1.6 ...

physics - A spring with a spring constant of 1.50 102 N/m is attached to a 1.2 ...

physics - When the frequency of a force applied to a system matches the natural...

Physics - A mass-spring system has b/m = ω0/2, where b is the damping ...

Physics - A coiled Hookean spring is stretched 10 cm when a 1.5kg mass is hung ...

Theory of Vibration - (1) The time for vibration of a viscous-damped spring mass...

physics - The frequency of vibration of an object-spring system is 8.00 Hz when ...

physics - a 0.75 kg mass attached to avertical spring streches the spring 0.30m...