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Katarina [22]
3 years ago
5

A mass m on a spring with a spring constant k is in simple harmonic motion with a period T. If the same mass is hung from a spri

ng with a spring constant of 3k, the period of the motion will be _______.
a) increased by a factor of 3b) decreased by a factor of 3c) increased by a factor of the square root of 3d) decreased by a factor of the square root of 3
Physics
1 answer:
weqwewe [10]3 years ago
3 0

Answer:

(d) deceased by a factor of the square root of 3.

Explanation:

The period ,mass and spring constant of a spring in simple harmonic motion is related by the formula below

T=2\pi \frac{\sqrt{m} }{k}

where:

  • T is the period
  • m is mass
  • k is spring constant

for the first instance we will be using the subscript 1,

T_{1} = 2\pi \frac{\sqrt{m} }{k} ...............................equation 1

for the second instance we will use subscript 2. The spring constant is trippled (3k) while the mass remains same.

T_{2} =2\pi \frac{\sqrt{m} }{3k} ...........................equation 2

divide equation 2 by 1

\frac{T_{2} }{T_{1}} = \frac{2\pi \frac{\sqrt{m} }{3k} }{2\pi \frac{\sqrt{m} }{k}}

\frac{T_{2} }{T_{1}} = \frac{\sqrt{m} }{3k} ÷ \frac{\sqrt{m} }{k}

\frac{T_{2} }{T_{1}} = \frac{\sqrt{m} }{3k} × \frac{\sqrt{k} }{m}

\frac{T_{2} }{T_{1}} = \frac{\sqrt{k} }{\sqrt{3k} }

\frac{T_{2} }{T_{1}= \frac{\sqrt{k} }{\sqrt{3}\sqrt{k}  }

\frac{T_{2} }{T_{1}= \frac{1}{\sqrt{3} }

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Answer:

g' = 10.12m/s^2

Explanation:

In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.

You use the following formula:

T=2\pi \sqrt{\frac{l}{g}}         (1)

l: length of the pendulum = ?

g: acceleration due to gravity at sea level = 9.79m/s^2

T: period of the pendulum at sea level = 1.2s

You solve for l in the equation (1):

l=\frac{gT^2}{4\pi^2}\\\\l=\frac{(9.79m/s^2)(1.2s)^2}{4\pi^2}=0.35m

Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:

T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}

g': acceleration due to gravity at the top of the mountain

T': new period of the pendulum

g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}

The acceleration due to gravity at the top of the mountain is 10.12m/s^2

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A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f
Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

T=\dfrac{2\pi}{\omega}

{\omega}=\dfrac{2\pi }{0.2}\ rad/s

{\omega}=31.41\ rad/s

We know that

{\omega}^2=m\ K

K=Spring constant

K=\dfrac{\omega^2}{m}

K=\dfrac{31.41^2}{0.12}\ N/m

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

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