Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader-a physicist, naturally-immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.44 m. She sets the pendulum swinging, and her collaborators carefully count 107 complete cycles of oscillation during 2.00 x 10 s. What is the result? 

acceleration due to gravity: _______ m/s²

Answer 1

Answer 2

The pendulum's period is

T=tn" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 18px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">$$T=\frac{t}{n}$$

when n cycles of oscillation are counted during time t. For a simple pendulum,

T=2⁢π⁢LgPP" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 18px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">$$T=2\mathit{\pi }\sqrt{\frac{L}{g_{\text{PP}}}}$$

where L denotes its length and g_PP is the acceleration due to gravity on the surface of the Pink Planet. Square this equation and solve for g_PP.

gPP=L⁢(2⁢πT)2" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 18px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">$$g_{\text{PP}}=L{\left(\frac{2\mathit{\pi }}{T}\right)}^2$$

Substitute the expression for the period that was previously shown.

gPP=L⁢(2⁢π⁢nt)2" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 18px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">$$g_{\text{PP}}=L{\left(\frac{2\mathit{\pi }n}{t}\right)}^2$$

Now substitute the given numerical values.

gPP=(1.17 m)⁢(2⁢π×(100 oscillations)203 s)2=11.2⁢ m/s2" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 18px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?PP=(1.17 m)(2?×(100 oscillations)203 s)2=11.2 m/s2