Question

A mass-spring system with 0.200 kg undergoes simple harmonic motion with period 0.55 seconds. When an additional mass Δm is added, the period increases by 20 % . FIND Δm

show work

Answer 1

for a simple Harmonic motion,

T = 2m / k                                  { eq.1 }

Using above Equation for each case and Dividing them,

T₁ / T₂ = m / m + m                         { eq. 2 }

where, T₁ = initial time period = 0.55 sec

T₂ = final time period = (1 + 0.2) x 0.55 sec = 0.66 sec

m = mass of spring = 0.2 kg

m = change in mass = ?

inserting all these values in eq.2,

(0.55 sec) / (0.66 sec) = (0.2 kg) / (0.2 kg) + m

(0.833 sec) = (0.2 kg) / (0.2 kg) + m

squaring on both side, we get

(0.693889 sec) = (0.2 kg) / (0.2 kg) + m

(0.2 kg) + m = (0.2 kg) / (0.693889 sec)

(0.2 kg) + m = 0.288 kg.sec

m = (0.288 kg.sec) - (0.2 kg)

m = 0.088