Question

A 50-cm-long spring is suspended from the ceiling. A 290 g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 30 cm before coming to rest at its lowest point. It then continues to oscillate vertically.

(a) What is the spring constant? (K=)
(b) What is the amplitude of the oscillation?
(c) What is the frequency of the oscillation?

Answer 1

Answer:

a) Spring Constant $K=18.946\ \rm N/m$

b) Amplitude of the oscillation$A=0.3\ \rm m$

c)Frequency o the oscillation$\nu =1.287\ \rm rev/s$

Explanation:

Given:

• Magnitude of the mass, m=0.29 kg

• Spring is stretched by x=30 cm

a)

When Finally the mass comes to rest after 30 cm of stretching in spring then work done by all the forces is equal to the change in kinetic Energy of the of the mass

$\dfrac{Kx^2}{2}=mgh\\\dfrac{K\times0.3^2}{2}=0.29\times 9.8\times0.30\\K=18.946\ \rm N/m$

b) The Amplitude of the oscillation is 0.3 m

c) Let Frequency of the oscillation of the mass be$\nu$ which is given by

$\nu=\sqrt{\dfrac{K}{m}}\\\nu=\dfrac{1}{2\pi}\sqrt{\dfrac{18.946}{0.29}}\\\nu=1.287\ \rm rev/s$

Answer 2

a) Spring Constant     $K=18.946\dfrac{N}{m}$

b) Amplitude of the oscillation  $A=0.3m$

c)Frequency o the oscillation    $w=1.287 \dfrac{Rev}{sec}$

What will be the spring constant, Amplitude, and frequency?

It is given that Given:

mass Of an object, m=0.29 kg

Spring is stretched by x=30 cm

(1) While the spring is stretched then all PE will be converted to the KE of the mass

$\dfrac{1}{2} kx^2 =mgh$

$K\times 0.3^2 =2\times 0.29\times 9.8\times 0.30$

$K=18.946 \ \frac{N}{m}$

(2) The amplitude will be A=0.3m

(3) The frequency of the spring will be

$w=\dfrac{1}{2\pi }\sqrt{ \dfrac{k}{m} }$

By putting the values

$w= \dfrac{1}{2\pi } \sqrt{\dfrac{18.946}{0.29} } =1.287 \ \dfrac{rev}{sec}$

Thus  

a) Spring Constant     $K=18.946\dfrac{N}{m}$

b) Amplitude of the oscillation  $A=0.3m$

c)Frequency o the oscillation    $w=1.287 \dfrac{Rev}{sec}$

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