Answer:
(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:
- T is the period
- m is mass
- k is spring constant
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} }
