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

Explanation:

Here we have

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + F_kx$

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

$v_{max1} =A\sqrt{\frac{k}{m} }$

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;

$v_{max2} =A\sqrt{\frac{4 \times k}{m} } = 2 \times A\sqrt{\frac{ k}{m} }$

Therefore, $v_{max2} = 2 \times v_{max1}$, that is the velocity will increase by a factor of 2.