Question

When a particular mass-spring system is started by stretching the spring 3 cm, one complete oscillation takes 4 seconds.

(a) If instead the initial stretch had been 5 cm, how long would it take for one complete oscillation?
Time for one oscillation = s

(b) If the mass were doubled, what would the period of the system be?
Period = s

Answer 1

When the initial stretch is 5 cm, it will also take 4 seconds for one complete oscillation. When the mass is doubled, the period of the system will be the same as before, multiplied by a factor of square root of 2.

Step-by-step explanation:

To find the time for one complete oscillation when the initial stretch is 5 cm, we can use the equation for the period of a mass-spring system:

Period = 2π√(m/k)

Where m is the mass and k is the spring constant. Since the spring constant remains the same, the equation becomes:

Period = 2π√(m)

Since the mass does not change, the period remains the same.

Therefore, when the initial stretch is 5 cm, it will also take 4 seconds for one complete oscillation.

When the mass is doubled, the equation for the period becomes:

Period = 2π√(2m)

Therefore, the period of the system will be the same as before, multiplied by a factor of √2.

Answer 2

A. 5.16 s.

B. 5.66 s.

Step-by-step explanation:

A.

For a simple harmonic motion,

T = 2pi (sqrt * (l/g))

Given:

L1 = 3 cm

T1 = 4 s

L2 = 5 cm

T2 = ?

4 = 2pi*sqrt(3/g)

g = 7.4

At, L2,

T2 = 2pi*sqrt(5/7.4)

= 5.16 s.

B.

M1 = M1

M2 = 2*M1

For a simple harmonic motion,

T = 2pi (sqrt * (m/k))

4 = 2pi (sqrt * (M1/k))

M1/k = 0.405

Inputting the above values,

T2 = 2pi (sqrt * (2*M1/k))

= 2pi (sqrt * (2 * 0.405))

= 5.66 s.