Respuesta :
Answer:
Part a)
T = 3.96 s
Part b)
T = 1.98 s
Part c)
T = 2.8 s
Explanation:
As we know that time period of spring block system is given as
[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]
T = 2.8 s
Part a)
If the mass of the block attached is doubled
then we will have
[tex]T' = 2\pi\sqrt{\frac{2m}{k}}[/tex]
[tex]T' = \sqrt2 T[/tex]
[tex]T' = 3.96 s[/tex]
Part b)
If the spring constant is doubled
then we have
[tex]T' = 2\pi\sqrt{\frac{m}{2k}}[/tex]
[tex]T' = \frac{T}{\sqrt2}[/tex]
[tex]T' = 1.98 s[/tex]
Part c)
If the amplitude is halved but mass and spring constant will remain the same
so here we know that time period does not depends on Amplitude
so we will have
[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]
T = 2.8 s
The period of  block oscillating on a spring depends on its mass and the spring constant.
- (a) The period if the block's mass is doubled is 3.96 seconds.
- (b) The period if the value of the spring constant is doubled is 1.98 seconds.
- (c) The period if the oscillation amplitude is halved while m and k are unchanged is 2.8 seconds.
What is time period of oscillating spring block?
The time period of oscillating spring block is time taken by it to complete one cycle of swing left to right and right to left.
It can be given as,
[tex]T=2\pi \sqrt{\dfrac{m}{k}}[/tex]
Here, [tex]g[/tex] is the gravitational force of Earth and [tex]k[/tex] is the spring constant.
Given information-
The period of the oscillating block is 2.8 s.
- (a) The period if the block's mass is doubled-
The mass is doubled in the case (a). Put the value of mass as 2m in the  above formula as,
[tex]T_a=2\pi \sqrt{\dfrac{2m}{k}}\\T_a=2\pi \sqrt{\dfrac{m}{k}}\times\sqrt{2}\\T_a=T\times\sqrt{2}\\T_a=2.8\times\sqrt{2}\\\\T_a=3.96\rm s[/tex]
Thus, the period of the block if the block's mass is doubled is 3.96 seconds.
- (b) The period if the value of the spring constant is doubled-
The spring constant is doubled in the case (b). Put the value of spring constant as 2k in the  above formula as,
[tex]T_a=2\pi \sqrt{\dfrac{m}{2k}}\\T_a=2\pi \sqrt{\dfrac{m}{k}}\times\dfrac{1}{\sqrt{2}}\\T_a=T\times\dfrac{1}{\sqrt{2}}\\T_a=2.8\times\dfrac{1}{\sqrt{2}}\\\\T_a=1.98\rm s[/tex]
Thus, the period of the block if the spring constant is doubled is 1.98 seconds.
- (c) The period if the oscillation amplitude is halved while m and k are unchanged-
The oscillation amplitude is halved while m and k are unchanged in the case (c). As the period does not depend on the amplitude. Thus period will be,
[tex]T_c=2\pi \sqrt{\dfrac{m}{k}}\\T_c=2\pi \sqrt{\dfrac{m}{k}}\\T_c=T\times\dfrac{1}{\sqrt{2}}\\T_c=2.8[/tex]
Thus, the period of the block if the period if the oscillation amplitude is halved while m and k are unchanged is 2.8 seconds.
Hence,
- (a) The period if the block's mass is doubled is 3.96 seconds.
- (b) The period if the value of the spring constant is doubled is 1.98 seconds.
- (c) The period if the oscillation amplitude is halved while m and k are unchanged is 2.8 seconds.
Learn more about the time period of oscillating spring block here;
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