Answer:
The frequency will be reduced by a factor of √2/2
Explanation:
Pls see attached file
The new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
Let the initial mass of block be m.
And new mass is, 4m.
The frequency of oscillating motion is defined as the number of complete oscillation made during the time interval of 1 second. The mathematical expression for the frequency of oscillation of block-spring system is given as,
[tex]f = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
Here,
k is the spring constant.
If the mass of block increased to 4m, then the new frequency of oscillation of spring will be,
[tex]f' = \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{4m}}\\\\\\f' =\dfrac{1}{2} \times \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m}}\\\\\\f' =\dfrac{1}{2} \times f[/tex]
Thus, we can conclude that the new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
Learn more about the frequency of oscillation here:
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