Question

A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A block of mass m is suspended from this balance, when displaced and released, it oscillates with a period 0.5 s. The value of m is

(Take g = 10 ms–2)

• A. 8 kg
• B. 12 kg
• C. 16 kg
• D. 20 kg

Answer 1

Answer: (C)

16 kg

Answer 2

The 20 cm length of the scalar reads upto 50 kg

$\therefore \mathrm { F } = \mathrm { mg } = ( 50 \mathrm {~kg} ) \left( 10 \mathrm {~ms} ^ { - 2 } \right) = 500 \mathrm {~N}$

And

$\therefore$ Spring constant, $\mathrm { k } = \frac { \mathrm { F } } { \mathrm { x } } = \frac { 500 \mathrm {~N} } { 0.2 \mathrm {~m} } = 2500 \mathrm { Nm } ^ { - 1 }$

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

Squaring both sides, we get

$\mathrm { T } ^ { 2 } = \frac { 4 \pi ^ { 2 } \mathrm {~m} } { \mathrm { k } }$

$\mathrm { m } = \frac { \mathrm { T } ^ { 2 } \mathrm { k } } { 4 \pi ^ { 2 } } = \frac { ( 0.5 \mathrm {~s} ) ^ { 2 } \times \left( 2500 \mathrm { Nm } ^ { - 1 } \right) } { 4 \times ( 3.14 ) ^ { 02 } } = 16 \mathrm {~kg}$