Question

A block of mass m attached to a horizontally mounted spring with spring constant k undergoes simple harmonic motion on a frictionless surface. How would the maximum speed of the block be affected if the spring constant was increased by a factor of 4 while holding the amplitude of oscillation constant? It would decrease by a factor of 14 . It would increase by a factor of 4. It would remain unchanged. It would increase by a factor of 2. It would decrease by a factor of 12 .

Answer 1

Question:

a. It would decrease by a factor of 14 .

b. It would increase by a factor of 4.

c. It would remain unchanged.

d. It would increase by a factor of 2.

e. It would decrease by a factor of 12 .

Answer:

The correct option is;

d. It would increase by a factor of 2

Step-by-step explanation:

Here we have

`(1)/(2)kx^2 = (1)/(2)mv^2 + F_kx`

The formula for maximum mass is given by the following relation;

`v_(max1) =A\sqrt{(k)/(m) }`

Where:

`v_(max)` = Maximum speed of mass on spring

k = Spring constant

m = Mass attached to the spring

A = Amplitude of the oscillation

Therefore, when k is increased by a factor of 4 we have;

`v_(max2) =A\sqrt{(4 * k)/(m) } = 2 * A\sqrt{( k)/(m) }`

Therefore,
`v_(max2) = 2 * v_(max1)`, that is the velocity will increase by a factor of 2.