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Question

Question: A mass M is suspended by two springs of force constants K­<sub>1</sub> and K<sub>2</sub> respectivel...

A mass M is suspended by two springs of force constants K­1 and K2 respectively as shown in the diagram. The total elongation (stretch) of the two springs is

A

MgK1+K2\frac { M g } { K _ { 1 } + K _ { 2 } }

B

Mg(K1+K2)K1K2\frac { M g \left( K _ { 1 } + K _ { 2 } \right) } { K _ { 1 } K _ { 2 } }

C

MgK1K2K1+K2\frac { M g K _ { 1 } K _ { 2 } } { K _ { 1 } + K _ { 2 } }

D

K1+K2K1K2Mg\frac { K _ { 1 } + K _ { 2 } } { K _ { 1 } K _ { 2 } M g }

Answer

Mg(K1+K2)K1K2\frac { M g \left( K _ { 1 } + K _ { 2 } \right) } { K _ { 1 } K _ { 2 } }

Explanation

Solution

For series combination

keq=k1k2k1+k2k _ { e q } = \frac { k _ { 1 } k _ { 2 } } { k _ { 1 } + k _ { 2 } }

x=mg(k1+k2)k1k2\Rightarrow x = \frac { m g \left( k _ { 1 } + k _ { 2 } \right) } { k _ { 1 } k _ { 2 } }