A 0.5 Kg mass is supported by a spring . The system is set in vibration at its natural frequency of 4 Hz with an amplitude of 5 Cm.Find the spring constant .

Use 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

I 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

The book answer appears to be wrong by a factor of pi

Thanks a lot

To find the spring constant, we can use the formula for the frequency of an oscillating system:

f = 1 / (2π) √(k / m)

where:
f = frequency of the system (in Hz)
k = spring constant (in N/m)
m = mass of the object (in kg)

In this case, we are given the frequency (f = 4 Hz) and the mass of the object (m = 0.5 kg). We need to solve for the spring constant (k).

First, let's convert the amplitude from centimeters (cm) to meters (m) since all the other units in the formula are in SI units:

Amplitude (in m) = 5 cm = 5 / 100 = 0.05 m

Next, we can rearrange the formula to solve for the spring constant:

k = (4π² m) / (Amplitude)²

Now we can substitute the given values into the formula:

k = (4π² * 0.5) / (0.05)²
= (4 * 3.14² * 0.5) / (0.05)²
≈ 98.8 N/m

Therefore, the spring constant is approximately 98.8 N/m.