# Oscillation/amplitude question

#### unhip_crayon

Heres the question Im having problems solving...

A 3.00kg block hangs from a spring with a spring constant of 200.N/m and is set into vertical oscillation. The block has a velocity of 0.900m/s upwards when the spring is stretched by 11.0 cm. Calculate the amplitude of the oscillation.

So heres what I got...

K=200
m=3.00
V=0.900
x=0.11m
Xm=?

solve for Xm using this equation

Am I doing it right?
Help ASAP
Thank You

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#### Shooting Star

Homework Helper
solve for Xm using this equation

Am I doing it right?
Help ASAP
Thank You
Yes, provided the stretch is measured from the equilibrium position in which the mass hangs after being put on the spring. In this problem, that seems to be implied.

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#### unhip_crayon

but the answer I am getting is incorrect. The correct answer is 11.6cm (yes i did convert m to cm from my final answer)

Thanks

#### Troels

A 300kg block hangs from a spring with a spring constant of 200.N/m and is set into vertical oscillation. The block has a velocity of 0.900m/s upwards when the spring is stretched by 11.0 cm. Calculate the amplitude of the oscillation.

So heres what I got...

K=200
m=3.00
V=0.900
x=0.11m
Xm=?
It may help to note that you have somehow transmuted 300 kg into 3 kg!

Using the correct mass, the amplitude of the oscilation is 1.16 m or 116 cm - which is what is was supposed to be apart from a factor of 10? Did you mess up the decimals here as well? (No hard feelings, that happens for the best of us)

#### unhip_crayon

It may help to note that you have somehow transmuted 300 kg into 3 kg!

Using the correct mass, the amplitude of the oscilation is 1.16 m or 116 cm - which is what is was supposed to be apart from a factor of 10? Did you mess up the decimals here as well? (No hard feelings, that happens for the best of us)

Oops...sorry. It is suppose to be 3.00kg. According to the book, yes, the correct answer should be 11.6cm. Do you think you can solve it for me? Ive got a midterm today at 11.30

Thanks

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#### Troels

Oops...sorry. It is suppose to be 3.00kg.
Are you sure it is not 300 *grams* = 0.3 kg? In that case I get 11.5 cm, which is the best so far - otherwise I get 15.6 cm

Do you think you can solve it for me?
The general symbolic solution is pretty obvious:

$$x_{\textrm{max}}=\sqrt{x^2+\frac{mv^2}{k}}$$
Which is also what you will get from a diffential-eq approach. So it is correct - It seems that you just need to get the numbers right.

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#### unhip_crayon

Are you sure it is not 300 *grams* = 0.3 kg? In that case I get 11.5 cm, which is the best so far - otherwise I get 15.6 cm

The general symbolic solution is pretty obvious:

$$x_{\textrm{max}}=\sqrt{x^2+\frac{mv^2}{k}}$$
Which is also what you will get from a diffential-eq approach. So it is correct - It seems that you just need to get the numbers right.
ok...so maybe the books answer is incorrect because I got the same answer as you
Thanks

#### pet

can I just check that this is the same sort of question?

"A mass at the end of a spring oscillates with a period of 2.8s. The mazimum displacement of the mass from its equilibrium position is 16cm.

a) what is the amplitude of the oscillations?
b) i) its angular frequency;
ii) its maxium acceleration.

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