Question

A tray of mass 12 kg is supported by two identical springs as shown in figure below. When the tray is pressed down slightly and then released, it executes SHM with time period of 1.5 s. What is the spring constant of each spring? When a block of mass m is placed on the tray, the period of SHM changes to 3.0 s. What is the mass of the block?

Answer 1

The time period of oscillations is given by,

T = 2π√{(m)/(k₁ + k₂)}

⇒ m = {T²(k₁ + k₂)}/{4π²}    ...(i)

Here, T = 1.5 s, k₁ = k₂ = K(let) and m = 12 kg

Hence, 12 = {(1.5)² (k + K)}/{4 x 9.87}

⇒ ={2 x 2.25 K}/{4 x 9.87} = 12

or, k = {12 x 4 x 9.87}/{2 x 2.25} = 1.05 x 10² Nm^-1

when additional mass of 'm' kg is placed on the tray of mass 12 kg, then 'm' = m + 12 kg and T = 3 sec

Using equation (i), we get

m + 12 = {T²(k₁ + k₂)}/{4π²}

= {T² x 2k}/{4π²}

= {(3)² x 2 x 1.05 x (10)²}/{4 x 9.87}

= 47.87

or, m = 47.87 - 12

= 35.87 kg