KCET · Physics · Oscillations
A piston is performing S.H.M. in the vertical direction with a frequency of \( 0.5 \mathrm{~Hz} \). A block of \( 10 \)
\( \mathrm{kg} \) is placed on the piston. The maximum amplitude of the system such that the block remains
in contact with the piston is
- A \( 1.5 \mathrm{~m} \)
- B \( 1 \mathrm{~m} \)
- C \( 0.1 \mathrm{~m} \)
- D \( 0.5 \mathrm{~m} \)
Answer & Solution
Correct Answer
(B) \( 1 \mathrm{~m} \)
Step-by-step Solution
Detailed explanation
(B)
\(\mathrm{f}=0.5 \mathrm{~Hz}\)
\(\omega=2 \pi \mathrm{f}=\pi\)
for block to remain in contact with the piston at amplitude position
weight A the block = Force due to oscillation.
mg \(=\) ma
mg \(=\mathrm{m}\left(\omega^{2} \mathrm{~A}\right)\)
\(\mathrm{A}=\frac{g}{\omega^{2}}=\frac{10}{\Pi^{2}}=1 \mathrm{~m}\)
\(\mathrm{f}=0.5 \mathrm{~Hz}\)
\(\omega=2 \pi \mathrm{f}=\pi\)
for block to remain in contact with the piston at amplitude position
weight A the block = Force due to oscillation.
mg \(=\) ma
mg \(=\mathrm{m}\left(\omega^{2} \mathrm{~A}\right)\)
\(\mathrm{A}=\frac{g}{\omega^{2}}=\frac{10}{\Pi^{2}}=1 \mathrm{~m}\)
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