Amass m on a spring with a spring constant k is in simple harmonic motion with a period t. if the same mass is hung from a spring with a spring constant of 3k, the period of the motion will be
a) increased by a factor of 3b) decreased by a factor of 3c) increased by a factor of the square root of 3d) decreased by a factor of the square root of 3

(d) deceased by a factor of the square root of 3.

Explanation:

The period ,mass and spring constant of a spring in simple harmonic motion is related by the formula below

T=2\pi \frac{\sqrt{m} }{k}

where:

for the first instance we will be using the subscript 1,

T_{1} = 2\pi \frac{\sqrt{m} }{k} equation 1

for the second instance we will use subscript 2. The spring constant is trippled (3k) while the mass remains same.

T_{2} =2\pi \frac{\sqrt{m} }{3k} equation 2

divide equation 2 by 1

\frac{T_{2} }{T_{1}} = \frac{2\pi \frac{\sqrt{m} }{3k} }{2\pi \frac{\sqrt{m} }{k}}

\frac{T_{2} }{T_{1}} = \frac{\sqrt{m} }{3k} ÷ \frac{\sqrt{m} }{k}

\frac{T_{2} }{T_{1}} = \frac{\sqrt{m} }{3k} × \frac{\sqrt{k} }{m}

\frac{T_{2} }{T_{1}} = \frac{\sqrt{k} }{\sqrt{3k} }

\frac{T_{2} }{T_{1}= \frac{\sqrt{k} }{\sqrt{3}\sqrt{k}  }

\frac{T_{2} }{T_{1}= \frac{1}{\sqrt{3} }

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation: