(a) What is the length of a simple pendulum that oscillates with a period of 2.4

Question

(a) What is the length of a simple pendulum that oscillates with a period of 2.4 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?

________ m LM=______m

(b) What mass would you need to suspend from a spring with a force constant of 20 N/m in order for the mass-spring system to oscillate with a period of 2.4 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2

mE=_________kg   mM=__________kg

________ m LM=______m

Explanation / Answer

T= 2pi * sq.root(l/g)

Where, T is time period, l is length of pendulum and g is accn due to gravity,

For earth,

1.2= 2pi * rt.(l/9.8)

L= (1.2/2pie)^2 * 9.8
Where g = 9.8 m/s2

L= 0.3578 metres (answer)

Now on mars,

1.2 = 2pie* rt.(l/g)

g on mars =3.7m/s2

L= (1.2/2pie)^2*3.7

L= 0.1350 metres (answer)

Note, value of pie =22/7 = 3.14

Now, relation of time period with spring constant and mass.

T=2pie sq.root (m/k)

Where m is mass attatched and k is spring constant.

1.2=2pie * rt.(m/10)

M= (1.2/2pie)^2 * 10

M= 0.3651 kg

And for mars,

M= (1.2/2pie)^2* 10

M= 0.3651 kg

So, If spring contant remains same on mars, then time period for any fixed mass will be same.

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