Question

A block attached to a spring with unknown spring constant oscilates with a period of 4.2 s. Parts A to D are independent questions, each referring to the initial situation. PartA What is the period if the mass is doubled? Express your answer to two significant figures and include the appropriate units. 7-1 Value Units Subrnit Part B what is the period ส the mass is halved? Express your answer to two significant figures and include the appropriate units. ㆁ四 TValue Units Submit Part C What is the period if the amplitude is doublad? Express your answer to two significant figures and include the approprlate unlts. T- 1 Value Units Submit Part D What is the period if the spring constant is doubled? Express your answer to two significant figures and include the appropriate units. 7-1 Value Submit

Answer 1

Answer:

Explanation:

A )

Th expression for time period for a spring -mass system is as follows

T = 2π$\sqrt{\frac{m}{k} }$

So if mass is doubled , Time period will be √2 or 1.414  times or 5.9 s

B )

If the mass is halved , time period becomes 1 / √2 times or .70 times or 3.0 s

C)

Time period does not depend upon the amplitude of oscillation . So in this case time period will be unchanged ie 4.2 s

D )

As per the formula above , if spring constant k is doubled ,  time period will be 1 / √2 times or .70 times or 3.0 s

Answer 2

Answer:

Explanation:

Time period, T = 4.2 s

The time period of a body in SHM with a spring is given by

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

where, m is the mass of the body and k be the spring constant.

(a) As the time period of the body is directly proportional to the square root of the mass of the body, so if the mass is doubled, the time period becomes √2 times the original time period.

So, T' = √2 T

T' = 1.414 x 4.2 = 5.9 second

(b) If mass is halved, the new time period becomes 1/√2 times the original time period.

T' = 4.2 / 1.414 = 3 second

(c) Time period does not depend on the amplitude, so the time period remains same , i.e., 4.2 s.

(d) As the time period is inversely proportional to the square root of teh spring constant, so as the spring constant is doubled, time period becomes 1/√2 times the original time period.

T' = 4.2 / 1.414 = 3 second