Question

The period of oscillation of a simple pendulum is $\mathrm { T } = 2 \pi \sqrt { \frac { \mathrm { L } } { \mathrm { g } } }$ . Measured value of L is 10 cm know to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 50 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g?

• A. (a) 2%
• A. (b) 3%
• A. (c) 4%
• A. (d) 5%

Answer 1

(d)

Sol. Here , $\mathrm { T } = 2 \pi \sqrt { \frac { \mathrm { L } } { \mathrm { g } } }$

Squaring both sides, we get

$T ^ { 2 } = \frac { 4 \pi ^ { 2 } L } { g }$ or $\frac { 4 \pi ^ { 2 } L } { T ^ { 2 } }$

The relative error in g is

$\frac { \Delta \mathrm { g } } { \mathrm { g } } = \frac { \Delta \mathrm { L } } { \mathrm { L } } + 2 \frac { \Delta \mathrm { T } } { \mathrm { T } }$

Here, $\mathrm { T } = \frac { \mathrm { t } } { \mathrm { n } }$ and $\Delta \mathrm { T } = \frac { \Delta \mathrm { t } } { \mathrm { n } }$

$\therefore$ $\frac { \Delta \mathrm { T } } { \mathrm { T } } = \frac { \Delta \mathrm { t } } { \mathrm { t } }$

The errors in both L and t are the least count errors

$\therefore \frac { \Delta \mathrm { g } } { \mathrm { g } } = \frac { 0.1 } { 10 } + 2 \left( \frac { 1 } { 50 } \right) = 0.01 + 0.04 = 0.05$

The percentage error in g is

$\frac { \Delta g } { g } \times 100 = \frac { \Delta L } { L } \times 100 + 2 \left( \frac { \Delta \mathrm { T } } { \mathrm { T } } \right) \times 100$ $= \left[ \frac { \Delta \mathrm { L } } { \mathrm { L } } + 2 \left( \frac { \Delta \mathrm { T } } { \mathrm { T } } \right) \right] \times 100 = 0.05 \times 100 = 5 \%$