A spring fixed to a wall is attached to a 6 kg block of wood on the other end. The block rests on a frictionless surface. The spring with spring constant 100 N/m is compressed by 0.1 m and releases. With what frequency and period will the block/spring system oscillate?

-I drew the diagram, and the givens: K=100 N/m, F=0.1 m, and you're trying to find the frequency (f) and Period (T). But the equations to find both of those are reciprocal… f=1/T T=1/f, so I don't know how to find them.. Please help!

To find the frequency and period of the block/spring system, you can use the equation:

ω = √(k/m)

where ω is the angular frequency, k is the spring constant, and m is the mass of the block.

The angular frequency is related to the frequency (f) and the period (T) by the equations:

ω = 2πf
T = 1/f

Given that the spring constant, k, is 100 N/m and the mass, m, is 6 kg, we can calculate the angular frequency:

ω = √(k/m)
ω = √(100 N/m / 6 kg)
ω ≈ 5.77 rad/s

Now, we can calculate the frequency:

ω = 2πf
f = ω / (2π)
f ≈ 5.77 rad/s / (2π)
f ≈ 0.92 Hz

Finally, we can calculate the period:

T = 1/f
T = 1 / 0.92 Hz
T ≈ 1.09 s

Therefore, the block/spring system will oscillate with a frequency of approximately 0.92 Hz and a period of approximately 1.09 seconds.

To find the frequency and period of the oscillation of the block/spring system, you can use the equation:

Angular frequency (ω) = √(spring constant/mass)

Given that the spring constant (k) is 100 N/m and the mass (m) is 6 kg, you can calculate the angular frequency as follows:

ω = √(100 N/m / 6 kg)
= √(16.67 rad/s^2)

Frequency (f) is related to angular frequency (ω) by the equation:

f = ω / (2π)

Therefore, you can find the frequency by substituting the value of ω into the equation:

f = √(16.67 rad/s^2) / (2π)
≈ 0.838 Hz

Now that you have the frequency, you can calculate the period (T) using the equation:

T = 1/f

Substituting the value of f into the equation, you can find the period as follows:

T = 1 / 0.838 Hz
≈ 1.19 s

Therefore, the block/spring system will oscillate with a frequency of approximately 0.838 Hz and a period of approximately 1.19 s.